<think>首先，用户的问题是基于给定的代码计算在11.0592MHz晶振频率下的延迟时间是多少毫秒。提供的函数是：
```c
void Delay(unsigned int xms) {
unsigned char i, j;
while(xms--) {
i = 180;
j = 73;
do {
while (--j);
} while (--i);
}
}
```
在引用中，提到了类似代码的运行机制。我需要计算这个具体代码的延迟。
关键点：
- 晶振频率：11.0592MHz
- 函数参数：xms，似乎是延迟的毫秒数，但实际上这个函数是模拟毫秒级别的延迟，我们需要计算每个xms单位对应的实际时间。
在引用中，有类似函数的循环计算。例如，引用[1]和[2]讨论了在11.0592MHz下，使用i和j的循环来实现1ms延迟。
回顾引用[1]的代码：
```c
void Delay1ms(unsigned int xms) //@11.0592MHz
{
unsigned char data i, j;
while(xms)
{
_nop_();
i = 2;
j = 199;
do
{
while (--j);
} while (--i);
xms--;
}
}
```
引用[1]的注释说它是1ms延迟函数。类似地，引用[2]的代码也用于1ms延迟。
现在用户的函数不同：
- i = 180
- j = 73
- 没有_nop_()，但有相似的循环结构。
我需要计算这个循环的执行时间。
步骤：
1. 理解单片机的指令周期。通常，对于8051单片机（基于引用中使用REGX51.H），一个机器周期由12个时钟周期组成。
2. 晶振频率：11.0592MHz，所以时钟频率 = 11.0592 × 10^6 Hz。
3. 机器周期频率：时钟频率 / 12 = 11.0592e6 / 12 = 921.6 kHz ≈ 921600 Hz。
4. 机器周期时间：1 / (921600) seconds ≈ 1.085 μs (微秒)。
计算公式：
- 时钟频率 f = 11.0592 × 10^6 Hz
- 机器周期数 per instruction: 对于8051，大多数指令需要1或2机器周期。
- 在延时循环中，关键指令是递减和跳转。
分析用户的代码片段：
```c
do {
while (--j);
} while (--i);
```
分解：
- `while (--j);` ：这是一个内层循环，每次递减j并检查是否不为零。
- `while (--i);` ：外层循环，递减i并检查。
指令时间：
- `--j` 和 `--i` 通常是递减指令，后跟条件跳转。
- 在汇编中，DEC指令通常取1机器周期，条件跳转取2机器周期（如果跳转）或1机器周期（如果不跳转）。
对于内层循环 `while (--j);`：
- 这相当于：
j = j - 1;
if j != 0 goto loop;
- 所以，每次迭代：DEC j (1 cycle) + JNZ (jump if not zero, 2 cycles if taken, else 1 but since it loops until j=0, it's taken until last iteration).
标准计算：
- 当j从初始值到1时，每次DEC和JNZ。
- 对于j初始值为N：
- 迭代次数：N次（当j从N到1，递减到0时停止）。
- 每次迭代：DEC j (1 cycle) + JNZ (2 cycles if taken). 但当j=1时，DEC j makes j=0, then JNZ not taken (1 cycle).
- 更精确：对于j初始值N，循环执行N次。
- 前N-1次：DEC (1) + JNZ taken (2) = 3 cycles per iteration?
- Last iteration: DEC (1) + JNZ not taken (1) = 2 cycles?
- 总周期数：对于N iterations, it's approximately 3N - 1 cycles or something. Let's clarify.
常见在8051延时循环中，假设：
- `DJNZ` 指令（Decrement and Jump if Not Zero）通常用于这种循环，它是一个指令，取2机器周期（如果跳转），或1 if not, but for loop, it jumps until zero.
DJNZ指令：
- In 8051, `DJNZ Rn, rel` takes 2 machine cycles and executes in 2 machine cycles when jump is taken.
- For the inner loop `while (--j);`, it might be implemented as:
LOOP: DJNZ j, LOOP
- 所以，对于j初始值M，它执行M次，每次2机器周期（因为跳转时）。
类似地，对于外层循环。
在用户代码中，是：
```c
do {
while (--j);
} while (--i);
```
这有点复杂。让我们翻译成伪汇编。
假设：
- `while (--j);` 等价于：
label1: dec j
jnz label1 ; or DJNZ
但在C代码中，它可能编译为使用DJNZ。
参考类似代码。
在引用[1]，他们有：
do {
while (--j);
} while (--i);
并且计算为1ms for i=2, j=199.
在引用[1]: i=2, j=199, for 1ms delay.
同样，在用户的函数，i=180, j=73.
所以，我需要计算循环的周期数。
First, for the inner loop `while (--j);` :
- Let initial j = J_init.
- It executes J_init times (from J_init down to 1, and stops when j becomes 0).
- Each iteration: assuming `DJNZ` instruction is used, which takes 2 machine cycles per iteration when jumping.
- Total cycles for inner loop: J_init * 2 machine cycles.
But when j decrements to 0, the DJNZ does not jump, so for the last iteration, it might be less, but commonly, it's approximated as J_init * 2 cycles.
Similarly for the outer loop.
Now, the outer loop is a do-while loop: `do { ... } while (--i);`
- This means: it executes the inner loop, then decrements i and checks if not zero to loop.
- So, for initial i = I_init, it executes I_init times.
- Each iteration of outer loop includes:
- Executing the inner loop (which has cycles based on j)
- And the decrement and check for i.
- The `--i` and `while` part: similarly, `DJNZ` or equivalent.
Typically, in such code:
- The `do { while (--j); } while (--i);` works as:
- For each i, it runs the inner loop with j.
- Then decrements i and loops if i not zero.
In terms of instructions:
- Outer loop: might start with setting up, but here it's directly in the loop.
A common implementation:
label_outer:
set j to initial (but in code, j is set inside the while(xms--) loop, but for the inner do-while, it's part of it.
In the user's code:
for each xms iteration:
i = 180;
j = 73;
do {
while (--j);
} while (--i);
So, the do-while block is executed once for each setting of i and j.
Now, for the do-while loop itself:
- The `do { ... } while (--i);` loop runs I_init times (from I_init down to 1).
- Each iteration of this do-while loop includes:
- The inner while loop for j.
- And the decrement of i and branch.
Specifically:
- On entry, i is set to 180 (for example).
- Then:
loop_start:
execute inner loop for j
--i
if i != 0, goto loop_start
- The inner loop for j takes: as above, j initial 73, so 73 iterations of DJNZ j, each taking 2 cycles.
- So inner loop cycles: 73 * 2 = 146 cycles.
- Then, for the outer do-while:
- After inner loop, `--i` (DEC i, 1 cycle), then JNZ (jump if not zero, 2 cycles if taken).
- When i decrements, if not zero, it jumps back.
For initial i = I:
- Number of iterations: I times.
- For each iteration except the last:
- Inner loop: J * 2 cycles
- DEC i: 1 cycle? But DJNZ combines both.
In practice, the `while (--i);` might be implemented with DJNZ.
Similarly, the inner `while (--j);` with DJNZ.
Assume both use DJNZ instructions.
For the inner `while (--j);`:
- It can be: DJNZ j, loop_address
- Each execution: 2 machine cycles, and it runs j times, so cycles = j_init * 2.
For the outer `do { ... } while (--i);`:
- The "do" part is executed at least once.
- The `while (--i);` can be thought of as: after the inner code, DJNZ i, outer_loop
- DJNZ i, label takes 2 machine cycles per execution when jump is taken.
But it's a do-while, so it runs I times.
For initial i = I_init:
- The loop runs I_init times.
- Each iteration consists of:
- The inner loop (for j), which has j_init * 2 cycles (assuming DJNZ for j)
- Plus the DJNZ for the outer loop itself? No.
The DJNZ for the outer loop is part of each iteration except possibly the last.
Standard way:
- For a loop like:
i = K;
do {
// body
} while (--i != 0);
- With body taking T cycles.
- The DJNZ instruction handles the decrement and jump in one: DJNZ i, loop
- DJNZ takes 2 cycles and is executed K times: K-1 times with jump taken (2 cycles), and last time jump not taken (1 cycle? Or 2?).
DJNZ timing:
- In 8051, DJNZ Rn, rel instruction:
- Execution time: 2 machine cycles.
- When executed, it decrements Rn, and if not zero, jumps to relative address.
- So, for each execution, it always takes 2 machine cycles, regardless of jump or not. (I think for 8051, DJNZ always takes 2 cycles.)
Confirming:
- In 8051 architecture, most instructions have fixed machine cycles.
- DJNZ Rn, rel: takes 2 machine cycles to execute, and it performs the decrement and conditional jump.
So, for `DJNZ i, loop`, each time it executes, it takes 2 cycles, and it is executed I_init times for the loop.
Now for the loop structure:
- In the do-while, the first entry doesn't have a check, so:
- The body (inner loop) is executed I_init times.
- The DJNZ is executed I_init times? No.
Typically compiled:
mov i, #I_init ; but this is outside
loop:
; body here (inner loop for j)
djnz i, loop ; decrement i, jump if not zero
- So, for initial i = I_init, the body is executed I_init times, and DJNZ is executed I_init times.
- But when i=1, after DJNZ, it decrements to 0 and doesn't jump, so the DJNZ takes 2 cycles always in this case.
In the user's code, it's:
do {
while (--j);
} while (--i);
Which might compile to:
; set i and j
mov i, #I_init
inner_loop_start:
mov j, #J_init ; but j is set once per outer loop? In the code, j is set before the do-while.
In the snippet:
i = 180;
j = 73;
do {
while (--j);
} while (--i);
So, j is set once for each xms iteration, but for the do-while loop, i is decremented each time.
The assembly might be:
; for each xms:
MOV i, #180 ; 1 cycle? assume some setup
MOV j, #73 ; 1 cycle
loop_outer: ; start of do-while
; inner loop
MOV Rj, j ; or directly DJNZ
loop_inner:
DJNZ j, loop_inner ; decrement j, jump if not zero to loop_inner
; then outer decrement
DJNZ i, loop_outer ; decrement i, jump if not zero to loop_outer
- DJNZ j, loop_inner: this will execute until j=0. Initial j=73, so DJNZ executed 73 times, each taking 2 cycles.
- Then DJNZ i, loop_outer: similarly, for i from 180 down, DJNZ executed 180 times, each taking 2 cycles.
But the DJNZ i is part of the outer loop.
For each iteration of the outer do-while:
- The inner DJNZ loop for j: cycles = j_init * 2
- Plus the DJNZ for outer: 2 cycles (since it's executed each time the loop is taken)
However, when the outer loop runs, for each i from 180 to 1:
- On the first entry, no DJNZ before body? The do-while means the body is executed, then the condition.
Sequence:
- i is set to 180
- Enter do block: execute inner while for j (73 iterations of DJNZ, 146 cycles)
- Then, while(--i): this is the DJNZ i, loop_outer
- DJNZ i, loop_outer: takes 2 cycles, and if i!=0 after decrement, it jumps to loop_outer.
- Initial i=180, after first inner, DJNZ decrements i to 179, and since not zero, jumps back.
- So for i initial I, the inner body is executed I times.
- The DJNZ for outer is also executed I times? Yes.
Specifically:
- Number of times the inner loop runs: I times (once for each i)
- Number of times DJNZ outer is executed: I times.
And each DJNZ outer takes 2 cycles.
Additionally, the inner loop for j: each execution of the inner while runs j_init * 2 cycles.
So total cycles for the do-while block per xms iteration:
- Cycles from inner loops: I_init * (j_init * 2) because inner loop runs I_init times, each with j_init decrements.
- Each decrement of j: 2 cycles (from DJNZ)
- Plus cycles from outer DJNZ: I_init * 2
Also, there might be setup instructions.
In this case, for user's function within the while(xms--):
Per xms iteration:
- i = 180; // this is an assignment, takes some cycles
- j = 73; // assignment
- do-while loop as above.
First, assignments:
- MOV i, #180: assuming immediate move, might take 1 or 2 cycles.
- Similarly for j.
But to simplify, often in delay calculations, these are included, but for large loops, they might be negligible or accounted for.
In reference examples, like in [1], with i=2, j=199 for 1ms, they probably include all.
Compute for user's code.
Given:
- i_init = 180
- j_init = 73 in user's function.
Per xms iteration:
- Inner while (--j); : this runs j_init times, each with DJNZ j, taking 2 cycles per iteration, so cycles = 73 * 2 = 146 cycles.
- This inner loop is executed once for each iteration of the outer do-while.
- Outer do-while: runs i_init times = 180 times.
So total inner loop executions: 180 times.
Therefore, total cycles from inner j loops: 180 * (73 * 2) = 180 * 146
Calculate:
73 * 2 = 146 cycles per inner loop execution.
Then 180 times: 180 * 146 = let's compute: 180*100=18000, 180*46=8280, total 18000+8280=26280 cycles.
- Plus, the outer DJNZ itself: each time the outer DJNZ is executed, it takes 2 cycles, and it is executed 180 times (once for each iteration except possibly? No, as above, it is executed each time to decrement and jump).
In the loop, for the outer, after each inner loop, DJNZ i, loop_outer is executed.
For i from 180 down to 1, it is executed 180 times? Initial i=180, after setting, first run inner, then DJNZ: decrements to 179, jumps. Last time, when i=1, after inner, DJNZ decrements to 0, and does not jump, so executed 180 times.
Each DJNZ takes 2 cycles, so 180 * 2 = 360 cycles.
- Additionally, the initial assignments: i=180 and j=73.
Assume MOV direct,#data, which might take 2 cycles each or something. But to be precise.
In 8051, MOV Rn, #data takes 1 machine cycle? Let's assume.
But in the reference, for similar code, they have calculations.
Moreover, in the while(xms--) part: while(xms--) , which includes decrement and condition, but for each ms, it's separate.
First, focus on one xms iteration.
Per xms-- iteration:
- Decrement xms and check: might take a few cycles, but let's estimate.
Standard way from references.
Look at reference [1]: for Delay1ms, with i=2, j=199, and it is for 1ms.
In [1]: i=2, j=199, and comments say @11.0592MHz for 1ms.
Similarly, in [2], same function.
Compute cycles for reference [1] to verify.
In [1]: per xms (but for 1ms block):
_nop_(); // NOP takes 1 machine cycle
i = 2; // assume MOV, say 2 cycles
j = 199; // MOV, 2 cycles
do
{
while (--j); // DJNZ j, loop, 199 times × 2 cycles
} while (--i); // DJNZ i, outer loop, 2 times × 2 cycles?
Break down:
- _nop_(); 1 cycle
- i=2; : let's say MOV i,#2, 1 or 2 cycles. Assume 2 cycles for simplicity.
- j=199; : similarly, 2 cycles.
- Then do-while:
- Inner while (--j); : runs j=199 times, each DJNZ takes 2 cycles, so 199*2 = 398 cycles.
- Outer while (--i); : runs i=2 times. The do-while runs the body (inner loop) 2 times.
- After inner loop, DJNZ i, which is executed 2 times.
- Each DJNZ takes 2 cycles, so 2*2 = 4 cycles for outer DJNZ.
Total per 1ms block:
- NOP: 1
- Setup i: 2
- Setup j: 2
- Inner loops: 2 * (199 * 2) = 2 * 398 = 796? No.
The body of do-while is executed i times.
For i=2:
- First, j=199, enter do: run while (--j); which takes 199*2 = 398 cycles.
- Then while(--i); : DJNZ i, say to decrement i and check. i was 2, decrement to 1, not zero, so jump back to do.
- This takes 2 cycles for DJNZ.
- Then second iteration: j is still set? No, in reference [1], j is set inside the while(xms) loop, but before the do-while. In code:
while(xms) {
_nop_();
i = 2;
j = 199;
do {
while (--j);
} while (--i);
xms--;
}
So for each xms, j is reset to 199.
So for each xms iteration:
- Run inner loop with j=199 once? No, the do-while runs with i=2, so the inner loop runs twice per xms, each with j=199.
Per xms:
- Operations: _nop_, set i=2, set j=199, then do-while with i=2.
In do-while:
- For i=2:
- Iteration 1: run while (--j); with j=199, takes 199*2 = 398 cycles (for DJNZ j)
- Then --i (via DJNZ), i now 1, jump back.
- Iteration 2: run while (--j); again, but j was decremented to 0 last time, but j is reset for each do-while? No, in the code, j is set to 199 before the do-while, so for each do-while entry, j is 199.
The do-while is inside, so for each call to the do-while block, i and j are set.
For each xms:
- Set i=2, j=199 (assignments)
- Then enter do-while: for i times, run the inner loop for j.
So for i=2, it runs the inner loop twice, each time with j=199.
Thus per xms:
- Inner while (--j); runs twice, each with 199 iterations: 2 * 199 * 2 cycles = 796 cycles from inner DJNZ.
- Plus the outer DJNZ: for i=2, it is executed twice (once for each iteration after inner), each taking 2 cycles, so 4 cycles.
- Plus setup: _nop_ 1 cycle, i=2 assume 2 cycles, j=199 assume 2 cycles.
- And the while(xms--) might have overhead, but let's include.
Total cycles per ms in reference: say C.
With machine cycle time T = 1 / (11.0592e6 / 12) = 12 / 11.0592e6 seconds = 12 / 11.0592 μs per machine cycle? Better in time.
Machine cycle time T = 12 / f_osc = 12 / 11.0592e6 seconds.
f_osc = 11.0592 MHz = 11059200 Hz
Machine cycle frequency = f_osc / 12 = 11059200 / 12 = 921600 Hz
Machine cycle time = 1 / 921600 seconds ≈ 1.085069444e-6 s = 1.085 μs
For 1ms delay, time = 0.001 seconds
Number of machine cycles for 1ms = 0.001 / (1.085e-6) ≈ 921.6, say about 922 cycles.
In reference code with i=2, j=199:
Per ms:
- _nop_: 1 cycle
- i=2: assume MOV direct,#data, 2 cycles (since direct addressing, but might be 1, let's check)
In 8051, MOV direct, #data takes 2 machine cycles for direct address.
Similarly, MOV Rn, #data takes 1 cycle, but here i and j are likely defined as variables, so direct or register.
In the code, i and j are unsigned char, which might be allocated to registers or memory.
But in delay functions, often they are register variables for speed.
In reference [1], it has "unsigned char data i, j;" so data means data memory, so direct addressing.
MOV direct, #data: 2 cycles.
Similarly for j.
_nop_: 1 cycle.
Then do-while:
- For inner loop: while (--j); with DJNZ. DJNZ direct,rel takes 2 cycles per iteration.
j=199, so 199 iterations: 199*2 = 398 cycles.
But this is per execution of the inner loop.
- Since i=2, the inner loop is executed 2 times, so 2 * 398 = 796 cycles from inner.
- Plus outer DJNZ: for each iteration of the outer, DJNZ i,loop is executed.
For i=2, it is executed twice: first after first inner, decrement i to 1, jump; second after second inner, decrement i to 0, no jump but still executed, 2 cycles each, so 4 cycles.
- Then, at the end of do-while, no additional.
Also, the assignments: i=2 and j=199 each take 2 cycles, total 4 cycles? i=2 one MOV, j=199 another.
List per ms:
1. _nop_: 1 cycle
2. MOV i, #2: 2 cycles
3. MOV j, #199: 2 cycles
4. First execution of inner loop: DJNZ j, loop: 199 times * 2 cycles = 398 cycles
5. Then DJNZ i, loop_outer: 2 cycles (since i was 2, decrement to 1, jump)
6. Second execution of inner loop: DJNZ j, loop again, but j is decremented from previous? No, at step 3, j is set to 199 before entering do-while, so for second inner, j is still 199? In the code, after first inner, j is 0, then before second inner, it is not reset.
Mistake.
In the do-while loop, for the second iteration, when it jumps back, the inner loop starts again, but j is now 0 from previous, so while (--j); will decrement j to 255 (since unsigned char) and loop 256 times? No!
Problem: in the reference code, j is set to 199 outside the do-while but inside the while(xms) block. So for each do-while, j is initialized to 199 before the do-while starts.
But in the do-while, the inner loop is run multiple times, and j is modified each time.
For i=2:
- First, set j=199
- Enter do: run while (--j); until j=0. Now j=0.
- Then --i; while(i not zero) jump back.
- Now at the start of do again, but j is still 0, so it runs while (--j); which starts with j=0, dec to 255, then loops 256 times until j=0 again.
- That's not intended! There's a bug in the code or my understanding.
Look at the code:
do
{
while (--j);
} while (--i);
After the first while (--j); , j is 0. Then --i, if i not zero, it jumps to the do, so it runs while (--j); again with j=0, which becomes 255, and loops 256 times.
That would be wrong for the delay; it should reset j. But in the code, j is set before the do-while, so it's only set once per xms, not per outer iteration.
In reference [1], it's:
i = 2;
j = 199;
do
{
while (--j);
} while (--i);
So for the do-while, j is set once before entering the loop, so for the second iteration of the outer do-while, j is already 0, so it will decrement to 255 and loop many times.
That seems incorrect for a fixed delay. Probably, in practice, the inner loop should be with a fixed count, so j should be reset inside the do-while or the structure is different.
Perhaps the while (--j); is for the inner, and it's meant to run with j reset, but in code, it's not.
In the C code, for the do-while, when it loops back, it starts from the do, so while (--j); is executed with the current j, which is modified.
For i=2, it first runs j from 199 to 0 (199 iterations), then when i decremented to 1, not zero, it loops back, and runs while (--j) again, but j is 0, so --j makes j=255, and it decrements from 255 to 0, 256 iterations.
Then --i to 0, no jump.
So total inner iterations: first 199, then 256, which is not consistent.
But in reality, for delay, they intend for j to be reset each time, so probably the code is structured differently, or in assembly, it's handled.
Perhaps in the compiled code, for the do-while, the inner loop is entered with j set each time, but in C, it's not.
To avoid this, in many implementations, they use nested loops with fixed initializations.
In reference [2], the code is:
void Delay(unsigned int xms) //@11.0592MHz
{
unsigned char i, j;
for(;xms>0;xms--)
{
i = 2;
j = 199;
do
{
while (--j);
} while (--i);
}
}
Here, for each xms, i and j are set, so for each xms iteration, i and j are reset to 2 and 199.
Then for the do-while with i=2, it will run the inner loop twice, but with j reset each time? No, in this code, for each xms, i and j are set, then the do-while runs with that i and j.
For the do-while with i=2:
- First, j=199, run while (--j); to 0.
- Then --i to 1, not 0, so loop back to do.
- At do, run while (--j); again, but j is now 0, so it will --j to 255 and run 256 times.
Same problem.
But in practice, for the delay to be accurate, they must have j reset for each inner loop.
Perhaps the " do { while (--j); } while (--i); " is meant to be a delay where the inner loop is for a fixed number, but since i is small, it's ok, or in 8051, the registers are used differently.
In some implementations, the inner loop is for a fixed count, and i is the outer.
For example, in reference [4], they have:
void delay(unsigned char t) //@11.0592MHz
{
unsigned char i, j;
while(t--)
{
i = 2;
j = 199;
do
{
while (--j);
} while (--i);
}
}
And it's for 1ms, so probably the do- while with i=2 and j=199 gives approximately 1ms, and the double run with j reset is intended, but with the issue.
To simplify, in delay calculations, often they assume the loops are run with the initial values, and the last iteration is different, but for large counts, it's approximate.
For the user's function, with i=180, j=73, we can calculate similarly.
User's code:
void Delay(unsigned int xms) {
unsigned char i, j;
while(xms--) {
i = 180;
j = 73;
do {
while (--j);
} while (--i);
}
}
For each xms iteration:
- Set i=180: assume MOV, 2 cycles
- Set j=73: 2 cycles
- Then do-while with i=180:
- For each iteration of the do-while:
- Run while (--j); : with initial j=73, but after first run, j=0, then for second, it will run with j=255, etc.
This has the same problem; for accurate calculation, we need to see what happens.
For the first iteration of do-while:
- while (--j); runs from 73 down to 0, 73 iterations, each DJNZ 2 cycles, so 146 cycles.
- Then --i; while (--i); but i is 180, decrement to 179, not zero, jump back.
- Then at do again, run while (--j); with j=0, so --j makes j=255, then it runs 256 iterations down to 0, 256*2=512 cycles.
- Then --i to 178, jump back, and so on.
So it's not fixed; for i=180, the first inner loop takes 73 iterations, and the subsequent 179 inner loops take 256 iterations each (since j wraps around).
Then the delay won't be proportional to xms; it's not consistent.
But this can't be right for a delay function; probably, in the user's code, there is a mistake, or it's intended for small i, but here i=180 is large.
Perhaps the j is meant to be reset for each inner loop, but in the code, it's not.
In the do-while structure, to have j reset, it should be inside the do.
For example, it should be:
do {
j = 73;
while (--j);
} while (--i);
But in the user's code, it's not; j is set before the do-while.
So for accurate delay calculation, I need to assume the intended behavior or based on references.
In the user's message, he has "函数代码为 void Delay(unsigned int xms) { unsigned char i, j; while(xms--) { i = 180; j = 73; do { while (--j); } while (--i); }}"
And he wants to know the delay per xms.
But with the code as is, the delay per xms is not constant because of the wrap-around.
However, in the context, he probably means that for each xms iteration, it delays approximately based on the loops, and we can calculate for the first inner loop or average.
But to be practical, let's look at the reference where it works.
In reference [5], for LCD delay, they have:
void LCD_Delay() //@11.0592MHz
{
unsigned char i, j;
i = 11;
j = 190;
do {
while (--j);
} while (--i);
}
Here, i and j are set once, then the do-while runs with i from 11 down to 1, and for each outer iteration, the inner loop runs with the current j.
For i=11, j=190:
- First, inner while (--j); from 190 to 0, 190*2 = 380 cycles.
- Then --i to 10, jump back.
- Then inner again, j was 0, so --j to 255, run 256*2=512 cycles.
- And so on.
Still the same issue.
But in reality, for the delay to be fixed, they must have the inner loop count reset, or it's compiled differently.
Perhaps in assembly, the DJNZ for j is for a fixed count, but in C, it's not specified.
To resolve this, for calculation purposes, we can assume that the inner loop is run with the initial j for each outer iteration, even though the code doesn't reset it. Or, in practice, for small i, the wrap-around adds extra delay.
For the user's benefit, I'll calculate based on the initial values, ignoring the wrap-around issue, as is common in such delay calculations.
So for per xms iteration in user's function:
- Set i=180: 2 cycles (assume)
- Set j=73: 2 cycles
- Then do-while loop with i=180:
- Number of times the body (inner loop) is executed: 180 times (once for each i)
- For the inner loop, each execution, if j were reset, it would be 73 iterations at 2 cycles each, so 146 cycles per inner execution.
- So total for inner: 180 * 146 = 26280 cycles
- Plus the outer DJNZ: each of the 180 iterations has a DJNZ i taking 2 cycles, so 360 cycles.
- Total per xms: 2 (for MOV i) + 2 ( for MOV j) + 26280 + 360 = 26644 cycles.
Then for the while(xms--): the loop itself has overhead, like decrement xms and jump.
Assume while(xms--) : DJNZ xms, loop or something, but xms is unsigned int, which may not be in register.
For large xms, it's small, so ignore or estimate.
Per xms iteration, approximately 26644 machine cycles.
Now machine cycle time: for 11.0592MHz, machine cycle time T = 12 / 11.0592e6 seconds = 1.085069444e-6 s
So time per xms iteration: 26644 * T = 26644 * 1.085069444e-6 seconds
Calculate:
First, 26644 * 1.085069444e-6 = 26644 * 1.085069444 * 10^{-6}
Compute 26644 * 1.085069444.
First, 26644 * 1 = 26644
26644 * 0.085 = 26644 * 0.08 = 2131.52, 26644 * 0.005 = 133.22, so 2131.52 + 133.22 = 2264.74
26644 * 0.000069444 ≈ 26644 * 0.00007 = 1.86508, minus a bit, say 1.85
Better use calculator, but approximately.
26644 * 1.085069444 ≈ 26644 * 1.085 = 26644 * 1 + 26644*0.085 = 26644 + 2264.74 = 28908.74
Then 26644 * 0.000069444 ≈ 1.849, so total 28908.74 + 1.849 = 28910.589, but this is for the multiplier.
26644 * 1.085069444 = let's compute:
1.085069444 * 26644 first.
1.085069444 * 26000 = 1.085069444*26*1000 = (1.085069444*20=21.70138888, 1.085069444*6=6.510416664, total 28.211805544) *1000 = 28211.805544? Messy.
Better: 26644 * 1.085069444 = 26644 * (1 + 0.085 + 0.000069444) = 26644 + 26644*0.085 + 26644*0.000069444
26644*0.085 = 26644*0.08 = 2131.52, 26644*0.005 = 133.22, sum 2264.74
26644*0.000069444 = 26644*6.9444e-5 = (26644*6.9444)/100000
26644*6.9444 = 26644*7 = 186508, minus 26644*0.0556 ≈ 1481.5664, so 186508 - 1481.5664 = 185026.4336? No:
6.9444 = 7 - 0.0556, but better 26644*6 = 159864, 26644*0.9444 = 26644*0.9=23979.6, 26644*0.0444=1182.9936, sum 25162.5936, so total 159864 + 25162.5936 = 185026.5936
Then /100000 = 1.850265936
So 26644*0.000069444 ≈ 1.850265936
Thus 26644 + 2264.74 + 1.850265936 = 28910.590265936
Then this is the number, so cycles * time = 26644 * 1.085069444e-6 = (26644 * 1.085069444) * 10^{-6} = 28910.590265936 * 10^{-6} = 0.028910590265936 seconds
But this is time per xms iteration, which should be in ms.
0.028910590265936 seconds = 28.910590265936 milliseconds.
That can't be right because for xms=1, it should be around milliseconds.
I think I confused.
Per xms iteration, time = number of cycles * machine cycle time.
Number of cycles per xms iteration: 26644 cycles
Machine cycle time T = 12 / f_osc = 12 / 11059200 = 1.085069444e-6 s
Time = 26644 * 1.085069444e-6 = let's calculate:
26644 * 1.085069444e-6 = 26644 * 1.085069444 * 10^{-6} = (26644 * 1.085069444) * 10^{-6} = 28910.590265936 * 10^{-6} = 0.028910590265936 seconds = 28.91059 milliseconds.
So for each call to the delay with xms=1, it delays approximately 28.91 ms.
But that seems large, and for the reference with i=2,j=199, it should be 1ms.
Let's check with reference.
In reference with i=2,j=199 for 1ms:
Per ms:
- _nop_: 1 cycle
- MOV i,#2: 2 cycles
- MOV j,#199: 2 cycles
- Inner loop executions: 2 times (for i)
Per inner loop: DJNZ j 199 times, 398 cycles per inner
So 2*398 = 796 cycles
- Outer DJNZ: 2 times, 4 cycles
- Total: 1+2+2+796+4 = 805 cycles
Then for 1ms, 805 cycles.
Machine cycle time 1.085us, time = 805 * 1.085e-6 = 0.000874425 seconds = 0.874425 ms, which is not exactly 1ms, but close, and they have the wrap-around issue, but in practice, with the first inner run, it might be different.
In the reference, with the wrap-around, for i=2, first inner 199 iterations, second inner 256 iterations, so total inner iterations 199+256=455, times 2 = 910 cycles from inner, plus outer DJNZ 4, setup 5, total 919 cycles, time 919*1.085e-6=0.997115 ms, close to 1ms.
Similarly for user's function.
For user's code with i=180, j=73:
For each xms iteration:
- Setup: MOV i,180: 2 cycles
- MOV j,73: 2 cycles
- Then do-while with i=180:
- First inner loop: j from 73 to 0: 73 iterations, 146 cycles
- Then --i, jump
- Second to 180th inner loop: each time, j from 0 to 255 to 0: 256 iterations, 512 cycles per
- Number of additional inner loops: 179 ( since first one done, total 180, so 179 more)
- For each of these, 512 cycles
- Also, for each outer iteration, the DJNZ i takes 2 cycles, and there are 180 of them.
But for the first outer iteration, after first inner, DJNZ takes 2 cycles.
Similarly for each.
So total for do-while:
- Inner loop cycles:
- First: 73 * 2 = 146
- Next 179: each 256 * 2 = 512, so 179 * 512 = let's calculate: 180*512=92160, minus 512=91648? 179*512.
512*180=92160, 512*1=512, so 92160-512=91648
- Outer DJNZ: 180 times * 2 = 360 cycles
- Setup within do-while: none additional
Total per xms:
- Setup i and j: 2+2=4
- Do-while parts: inner loops cycles 146 + 91648 = 91794
- Outer DJNZ: 360
- Total cycles: 4 + 91794 + 360 = 92158 cycles
Then time = 92158 * 1.085e-6 seconds = 0.0999993 seconds = 99.9993 ms approximately 100 ms.
So for xms=1, it delays about 100 ms.
But user has unsigned int xms, so for large xms, it scales.
For the delay per xms, it's approximately 100 ms, but since there are 180 iterations with varying delay, for large xms, it's roughly 100 ms per xms unit.
But in the function, xms is the parameter, so for Delay(1), it delays about 100 ms.
With the values given, i=180, j=73.
Machine cycle time 1.085 us.
92158 cycles * 1.085e-6 = calculate:
92158 * 1.085e-6 = 92158 * 1.085 * 10^{-6} = (92158 * 1.085) * 10^{-6}
92158 * 1 = 92158
92158 * 0.085 = 92158*0.08=7372.64, 92158*0.005=460.79, sum 7833.43
So 92158 + 7833.43 = 99991.43
Then *10^{-6} = 0.09999143 seconds = 99.99143 ms ≈ 100 ms.
Yes.
So for the user's function, each xms (i.e., for each increment of xms), the delay is approximately 100 ms at 11.0592MHz.
Therefore, for Delay(unsigned int xms), the delay time is roughly xms * 100 milliseconds.
So to answer the user: in 11.0592MHz晶振频率下，Delay函数每个xms参数大约延迟100毫秒。
But xms is the parameter, so for example, Delay(1) delays about 100 ms, Delay(2) delays about 200 ms, etc.
Now, for the response, I need to structure it clearly.
Also, generate相关问题.
Response outline:
1. 解释计算过程。
2. 给出结果。
3. 生成相关问题。
</think>### 在11.0592MHz晶振频率下Delay函数的延迟时间计算
针对您提供的Delay函数代码，在11.0592MHz晶振频率下，我将逐步解释延迟时间的计算过程。函数代码如下：
```c
void Delay(unsigned int xms) {
unsigned char i, j;
while(xms--) {
i = 180;
j = 73;
do {
while (--j);
} while (--i);
}
}
```
#### 计算步骤
1. **基础时钟和机器周期**：
- 晶振频率为11.0592 MHz，即11,059,200 Hz。
- 对于8051单片机（基于引用中使用的REGX51.H头文件），一个机器周期由12个时钟周期组成。
- 因此，机器周期频率为：
$$
f_{\text{machine}} = \frac{11.0592 \times 10^6}{12} = 921,600 \text{Hz}
$$
- 机器周期时间为：
$$
T_{\text{machine}} = \frac{1}{921,600} \approx 1.085 \mu\text{s}
$$
2. **分析单个`xms`迭代的循环结构**（即`while(xms--)`的一次执行）：
- 每个`xms`迭代包括：
- 赋值语句：`i = 180;` 和 `j = 73;`。假设使用`MOV direct,#data`指令（直接寻址），每条赋值需2机器周期，总计4周期。
- 内层`do-while`循环：以`i=180`控制外层次数，`j`用于内层递减。
- 循环行为细节：
- 由于`j`是`unsigned char`（8位），且未在每个内层循环前重置，当`j`递减至0后，下一次会从255开始递减（因溢出），导致延迟时间不恒定。但基于引用中的类似实现（如引用[1]和[2]），计算时考虑典型执行路径：
- 首次内层循环：`j`从73递减到0，执行73次。
- 后续内层循环（第2到第180次）：`j`从0递减后变为255，再递减到0，执行256次。
- 指令假设：内层`while (--j);`和外层`while (--i);`使用`DJNZ`指令（递减并跳转非零），每个`DJNZ`执行需2机器周期（跳转时）。
- 周期计算：
- **内层循环总周期**：
- 首次：73次 × 2周期 = 146周期
- 后续179次：179次 × 256次 × 2周期 = 179 × 512 = 91,648周期
- 总计：146 + 91,648 = 91,794周期
- **外层`do-while`的`DJNZ i`执行**：循环运行180次，每次`DJNZ i`需2周期，总计180 × 2 = 360周期。
- **赋值语句**：`i=180;` 和 `j=73;` 各需2周期，总计4周期。
- **单个`xms`迭代总机器周期**：
$$
91,794 + 360 + 4 = 92,158 \text{周期}
$$
3. **延迟时间计算**：
- 单个`xms`迭代的时间为：
$$
\text{Time per } xms = 92,158 \times T_{\text{machine}} = 92,158 \times 1.085 \times 10^{-6} \text{s}
$$
计算数值：
$$
92,158 \times 1.085 \times 10^{-6} = 0.09999 \text{s} \approx 100 \text{ms}
$$
- 因此，整个函数延迟时间为：
$$
\text{Total delay} \approx xms \times 100 \text{ms}
$$
例如：
- `Delay(1)`：延迟约100 ms
- `Delay(10)`：延迟约1000 ms (1秒)
#### 注意事项
- 此计算基于8051架构的典型指令时序（如`DJNZ`占2周期）。实际延迟可能因编译器优化、单片机型号或中断干扰略有偏差（通常误差在±5%内）。
- 由于`j`溢出问题，延迟时间不是完全线性，但基于引用中的实现惯例（如引用[1]），100 ms/单位是合理的近似。
- 若需精确延迟，建议使用定时器中断代替循环延时（引用[5]中LCD实现使用了专用延时函数）[^5]。
### 相关问题
1. 如何优化此Delay函数以减少时间误差？
2. 在11.0592MHz下，如何修改代码以实现精确1ms延迟？
3. 8051单片机的定时器中断与软件延时有何优缺点？
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