C 中的输出参数 [英] Out parameters in C

问题描述

void swap(int &first, int &second){
    int temp = first;
    first = second;
    second = temp;
}

int a=3,b=2;
swap(a,b);

编译器抱怨 void swap(int &first, int &second) 有语法错误.为什么?C 不支持引用吗?

The compiler complaints that void swap(int &first, int &second) has a syntax error. Why? Doesn't C support references?

推荐答案

C 不支持通过引用传递.因此，您需要使用指针来完成您想要实现的目标:

C doesn't support passing by reference. So you will need to use pointers to do what you are trying to achieve:

void swap(int *first, int *second){
    int temp = *first;
    *first = *second;
    *second = temp;
}


int a=3,b=2;
swap(&a,&b);

<小时>

我不推荐这个:但为了完整起见，我会添加它.

---

I do NOT recommend this: But I'll add it for completeness.

如果您的参数没有副作用，您可以使用宏.

You can use a macro if your parameters have no side-effects.

#define swap(a,b){   
    int _temp = (a); 
    (a) = _b;        
    (b) = _temp;     
}

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