(5)风色从零单排《C++ Primer》 const,typedef,auto,decltype

从零单排《C++ Primer》

——(5)const,typedef,auto,decltype  

CONST

多文件下

//file_1.cc defines and initializes a const that is accessible to other files
extern const int bufSize = fcn();

//file_1.h
extern const int bufSize;//same bufSize as defined in file_1.cc

Exercuse 2.26 Which of the following are legal?For those that are illegal,explain why.
(a)const int buf; //error must be initialized
(b)int cnt = 0; //ok
(c)const int sz = cnt; //ok
(d)++cnt;++sz; // error sz is read-only

const引用以及指针

Top-Level const

int i =0;
int *const p1 = &i; const int ci = 42;

Low-Level const

const int *p2 = &ci;//point to a const object
const int &r = ci;//const in reference types is always low-level

low与top levelconst之间的复制会受到限制

const int *const p3 = p2;

int *p = p3;//error:p3 has a low-level const but p doesn't

p2 = p3; //ok:p2 has the same low-level const qualification as p3

p2 = &i; // ok:we can convert int* to const int*

int &r = ci;//error:can;t bind an ordinary int& to a const int object

const int &r2 = i; // ok:can bind const int& to plain int

Exercise 2.30 : For each of the following declarations indicate whether the object being declared has top-level or low-level const.

const int v2 = 0; //top
int v1 = v2; //nonconst
int *p1 = &v1,&r1 = v1;//noneconst
const int *p2 = &v2, *const p3 = &i, &r2 = v2;//p2 is low-level,p3 is top-level, r2 is low-level

Exercise 2.31:
    int i = 0;
    const int v2 = 0; //top
    int v1 = v2; //nonconst
    int *p1 = &v1,&r1 = v1;//noneconst
    const int *p2 = &v2, *const p3 = &i, &r2 = v2;//p2 is low-level,p3 is top-level, r2 is low-level
    
    r1 = v2;//ok
    p1 = p2; //error v2 is top-level
    p2 = p1;//ok
    p1 = p3;   //error p3 is top-level
    p2 = p3;//ok

Constant 表达式以及 consterxpr

constant表达式必须满足以下两个条件：

1)不被改变

2)在编译时便可估算

const int max_files = 20;//constant expression
const int limit = max_files + 1; //limit is a constant expression
int staff_size = 27;//not a constant expression
const int sz = get_size();//not a constant expression

sz的值要在运行时才能初始化，为了解决这个问题，新标准引入了constexpr

constexpr int mf = 20; //20 is a constant expression
constexpr int limit = mf + 1;mf + 1 is a constant expression
constexpr int sz = size();//ok only if size is a constexpr function

constexpr int *q = nullptr;//q is a const pointer to int

typedef

使用typedef可以给类型一个新的名称，以方便使用。

typedef char *pstring;
const pstring cstr = 0;//cstr is a constant pointer to cahr
const pstring *ps;//ps is a pointer to a constant pointer to char

在新标准下，可以使用using

using SI = Sales_item;

auto

auto是一种推断类型。最多只能推断出low-level const。

auto i =0, *p = &i;// ok i is int and p is a pointer to int

auto sz = 0,pi = 3.14;//error inconsistent types for sz and pi

//Exercise 2.33 Using the variable definitions from this section,determine what happens in each of these assignmentss:
    int i =0, &r = i;
    auto a = r;
    const int ci = i, &cr = ci;
    auto b = ci; //top-level const in ci is dropped
    auto c = cr;
    auto d = &i;
    auto e = &ci;
    
    const auto f = ci;
    auto &g = ci;
    const auto &j = 42;
    
    a = 42; //ok
    b = 42; //ok
    c = 42; //ok
    //d = 42; error int to int *
    //e = 42; error int to int *
    //g = 42; error read-only

//Exercise 2.35 Determine the types deduced in each of the following definitions.Once you've figured out the types, write a program to see whether you were correct.
    
    const int i =42;
    auto j = i;//int
    const auto &k = i;//const int
    auto *p = &i;//const int *
    const auto j2 = i,&k2 = i;//const int, reference to const

decltype

decltype和auto累西，也是一种推断类型，但是它不需要像auto一样采取赋值初始化的方式来。可以推断出top-level const

decltype(f()) sum = x;// sum has whatever type f return
const int ci = 0;
decltype(ci) x = 0;//x has type const int

decltype((i)) d;// error remember that decltype((variable)) is always a reference type, but decltype(variable) is a reference type only if variable is a reference

decltype(i) e; // ok e is an (uninitialized) int

Exercise 2.36:
int a = 3, b = 4;
    decltype(a) c = a;// c is int
    decltype((b)) d = a;//d is reference
    ++c;
    ++d;

Exercise 2.37:

int a = 3, b = 4;
    decltype(a) c = a;
    decltype(a = b) d = a;//d也是引用，返回的事表达式左边的操作数的地址，也既是int &
    ++c;
    ++d;