标准输入输出

#include <iostream>

int main() {
    std::cout << "Enter two numbers:" << std::endl;
    int v1 = 0, v2 = 0;
    std::cin >> v1 >> v2;
    std::cout << "The sum of " << v1 << " and " << v2 << " is " << v1 + v2 << std::endl;
    return 0;
}

cin（see-in）：标准输入

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// 输入v1的值
int v1 = 0;
std::cin >> v1;

cout(see-out)：标准输出

1 2	// 输出字符串"Enter two numbers:" std::cout << "Enter two numbers:" << std::endl;

命名空间（namespace）

命名空间的出现是为了避免如下的问题：

我们声明了同名函数dance，那么如何调用指定的某个函数呢？

#include <iostream>

namespace Somebody {
    void dance() {
        std::cout<<"Somebody dances"<<std::endl;
    }
}

void dance() {
    std::cout<<"I dance"<<std::endl;
}

使用命名空间，指定调用的函数：

int main() {
    dance(); // 输出：I dance
    Somebody::dance(); // 输出: Somebody dances
    return 0;
}

引用（References）

引用就是一个对象的别名

A reference is not an object. Instead, a reference is just another name for an already existing object.

#include <iostream>
using namespace std;
int main() {
    int ival = 1024;
    int &refVal = ival;
    refVal = 512;
    cout << ival << endl; // 512
    int &refVal2; // error: references must be initialized.
    return 0;
}

引用必须初始化
一旦初始化，不可以再将引用指向其他对象

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int &refVal4 = 10; // error: initializer must be an object 
double dval = 3.14;
int &refVal5 = dval; // error: initializer must be an int object

Exercises Section 2.3.1

Exercise 2.15: Which of the following definitions, if any, are invalid? Why?

(a) int ival = 1.01;

(b) int &rval1 = 1.01;

(c) int &rval2 = ival;

(d) int &rval3;

(a) ✅（b）❌ 初始化赋值必须是int对象

指针（Pointers）

1 2	int ip1, ip2; // both ip1 and ip2 are pointers to int double dp, *dp2; // dp2 is a pointer to double; dp is a double

与引用不同，指针本身就是一个对象
在指针的生命周期中，可以指向不同的对象

1 2	int ival = 42; int *p = &ival; // p holds the address of ival; p is a pointer to ival

指针存放的值是另一个对象的地址

double dval;
double *pd = &dval; // ok: initializer is the address of a double
double *pd2 = pd; // ok: initializer is a pointer to double
int *pi = pd; // error: types of pi and pd differ
pi = &dval; // error: assigning the address of a double to a pointer to int

指针类型必须和其指向的地址中存放的对象的类型相同

注：& 和 * 有多种含义

&、*可以当做声明使用，比如：

int i = 42;
int&r=i; //& *followsatypeandispartofadeclaration; r is a *reference*

int *p; // * *follows a type and is part of a declaration; p is a *pointer*

&也可以作取地址符使用：

1	p = &i； // & is used in an expression as the address-of operator.

*可以作取值符号使用：

1 2	p = i; // is used in an expression as the dereference operator int &r2 = p; // & is part of the declaration; is the dereference operator

使用指针访问一个对象

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int ival = 42;
int *p = &ival; // p holds the address of ival; p is a pointer to ival 
cout << *p; // * yields the object to which p points; prints 42

使用指针指向空指针（Null Pointers）

int *p1 = nullptr; // equivalent to int *p1 = 0;
int *p2 = 0; // directly initializes p2 from the literal constant 0 
// must #include cstdlib
int *p3 = NULL; // equivalent to int *p3 = 0;

void指针（void* Pointers）

void指针能够指向任意类型的对象

double obj = 3.14, *pd = &obj;
// ok: void* can hold the address value of any data pointer type
void *pv = &obj; // obj can be an object of any type
pv = pd; // pv can hold a pointer to any type

Exercise 2.20: What does the following program do?

int i = 42;
int *p1 = &i;

*p1 = *p1 * *p1;

*p1 = 42 * 42

Leoolin

C++ Primer学习笔记

标准输入输出

命名空间（namespace）

引用（References）

指针（Pointers）

void指针（void* Pointers）