const pointer and array_ar[i]++-优快云博客

本文链接：https://blog.youkuaiyun.com/p_a_p_r_i_k_a/article/details/8642920

本文详细解析了C语言中常量指针的概念及应用，包括不同类型的指针定义方式及其特性，如只读整型变量、指向只读整型的指针等。同时对比了数组与指针的区别与联系，并解释了它们作为函数参数时的行为。

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const pointer starting from basic.

1.int var; //we get a int variable named var

2.const int var; //we get a int variable named var and it is read-only(const)

3.int *p; //from 1, we get a int variable *p(because * is the dereferencing operator, thus p is pointer)

4.const int *p; //from 2,we get a int variable *p and it is read-only

5.int *const p; //from 3,we get a int variable *const p, and p is read-only, this can be read as "int *(const p)"

6.const int *const p //from 4 and 5, we get a int variable *const p and it is read-only, p is also read-only. this can be read as "const int *(const p)"

array in C is actually compiled as pointer. if we define a array like int ar[n], then we can get its elements value by dereferencing it as a pointer *(ar + i).

but is array really a pointer? yes and no.

when defining an array : int ar[n], we actually define it like 5------int *const ar. when we defining a const arr : const ar[n], we define it like 6------const int *const ar

they are just the same!

when passed as parameter to a function, they act just the same. neither can you redirect it, nor change the value it points to. you can not even change the value it not point to by using it. suppose :

void foo(const int ar[]) {
    ar[i]++;                 //error
}

and

const int *const p;
(*(p  + 1))++;               //error too

however, you can change the value by using an other pointer

int *q = p;
(*(p + 1))++;                  //this is fine, you just get a warning

what an array can provide and a pointer can't is the size information : sizeof ar, but when passed to function, this information lost