在C OUT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);

在上面的例子中，C编译器抱怨作废掉期（INT和放大器;第一，INT和放大器;二）有一个像缺少一个语法错误＆放大器;之前（/ {我不明白，为什么？不支持C此功能，请澄清一下。谢谢。

In the above example, the C compiler complaints that "void swap(int &first, int &second)" has a syntax error like missing "&" before "( / {". I don't understand why? Doesn't C support this feature. Please clarify. Thanks.

推荐答案

C不支持引用传递。所以，你需要使用指针做你想达到什么样的：

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);

---

我的不的建议是：但我会添加它的完整性。

---

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

您可以使用宏，如果你的参数没有任何的副作用。

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

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

这篇关于在C OUT参数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！