C++引用分析实例与案例刨析及使用场景分析详解

int main() {

	int a = 10;
	int &b = a;

	cout << "a = " << a << endl;
	cout << "b = " << b << endl;

	b = 100;

	cout << "a = " << a << endl;
	cout << "b = " << b << endl;

	system("pause");

	return 0;
}

int main() {

	int a = 10;
	int b = 20;
	//int &c; //错误，引用必须初始化
	int &c = a; //一旦初始化后，就不可以更改
	c = b; //这是赋值操作，不是更改引用

	cout << "a = " << a << endl;
	cout << "b = " << b << endl;
	cout << "c = " << c << endl;

	system("pause");

	return 0;
}

//1. 值传递
void mySwap01(int a, int b) {
	int temp = a;
	a = b;
	b = temp;
}

//2. 地址传递
void mySwap02(int* a, int* b) {
	int temp = *a;
	*a = *b;
	*b = temp;
}

//3. 引用传递
void mySwap03(int& a, int& b) {
	int temp = a;
	a = b;
	b = temp;
}

int main() {

	int a = 10;
	int b = 20;

	mySwap01(a, b);
	cout << "a:" << a << " b:" << b << endl;

	mySwap02(&a, &b);
	cout << "a:" << a << " b:" << b << endl;

	mySwap03(a, b);
	cout << "a:" << a << " b:" << b << endl;

	system("pause");

	return 0;
}

void swap(int num1, int num2)
{
	cout << "交换前：" << endl;
	cout << "num1 = " << num1 << endl;
	cout << "num2 = " << num2 << endl;

	int temp = num1;
	num1 = num2;
	num2 = temp;

	cout << "交换后：" << endl;
	cout << "num1 = " << num1 << endl;
	cout << "num2 = " << num2 << endl;

	//return ; 当函数声明时候，不需要返回值，可以不写return
}

int main() {

	int a = 10;
	int b = 20;

	swap(a, b);

	cout << "mian中的 a = " << a << endl;
	cout << "mian中的 b = " << b << endl;

	system("pause");

	return 0;
}

//值传递
void swap1(int a ,int b)
{
	int temp = a;
	a = b; 
	b = temp;
}
//地址传递
void swap2(int * p1, int *p2)
{
	int temp = *p1;
	*p1 = *p2;
	*p2 = temp;
}

int main() {

	int a = 10;
	int b = 20;
	swap1(a, b); // 值传递不会改变实参

	swap2(&a, &b); //地址传递会改变实参

	cout << "a = " << a << endl;

	cout << "b = " << b << endl;

	system("pause");

	return 0;
}

//返回局部变量引用
int& test01() {
	int a = 10; //局部变量
	return a;
}

//返回静态变量引用
int& test02() {
	static int a = 20;
	return a;
}

int main() {

	//不能返回局部变量的引用
	int& ref = test01();
	cout << "ref = " << ref << endl;//第一次结果正确，编译器做了保留
	cout << "ref = " << ref << endl;//第二次可能会输出乱码

	//如果函数做左值，那么必须返回引用
	int& ref2 = test02();
	cout << "ref2 = " << ref2 << endl;
	cout << "ref2 = " << ref2 << endl;

	test02() = 1000;

	cout << "ref2 = " << ref2 << endl;
	cout << "ref2 = " << ref2 << endl;

	system("pause");

	return 0;
}

int * func()
{
	int a = 10;
	return &a;
}

int main() {

	int *p = func();

	cout << *p << endl;//第一次编译器可能会保留
	cout << *p << endl;//第二次直接会报错！取决于编译器。

	system("pause");

	return 0;
}

//发现是引用，转换为 int* const ref = &a;
void func(int& ref){
	ref = 100; // ref是引用，转换为*ref = 100
}
int main(){
	int a = 10;
    
    //自动转换为 int* const ref = &a; 指针常量是指针指向不可改，也说明为什么引用不可更改
	int& ref = a; 
	ref = 20; //内部发现ref是引用，自动帮我们转换为: *ref = 20;
    
	cout << "a:" << a << endl;
	cout << "ref:" << ref << endl;
    
	func(a);
	return 0;
}

//引用使用的场景，通常用来修饰形参
void showValue(const int& v) {
	//v += 10;
	cout << v << endl;
}

int main() {

	//int& ref = 10;  引用本身需要一个合法的内存空间，因此这行错误
	//加入const就可以了，编译器优化代码，int temp = 10; const int& ref = temp;
	const int& ref = 10;

	//ref = 100;  //加入const后不可以修改变量
	cout << ref << endl;

	//函数中利用常量引用防止误操作修改实参
	int a = 10;
	showValue(a);

	system("pause");

	return 0;
}