Question

C语言到底支不支持引用传递做函数参数？

同样一段代码，保存为cpp文件，在VC6.0下顺利运行，但是保存为c文件，连编译都不能通过。显示如下诸多错误：
error C2143: syntax error : missing ')' before '&'
error C2143: syntax error : missing '{' before '&'
......
代码如下：
#include<stdio.h>
typedef struct { int x, y,width,height; } Rect;
void Exchg3(Rect &x, Rect &y)
{
int tmp = x.height;
x.height = y.height;
y.height = tmp;
printf("x.height = %d,y.height = %d\n", x.height, y.height);
}
main()
{
Rect a = {1,2,3,4};
Rect b = {5,6,7,8};
Exchg3(a, b);
// printf("a = %d, b = %d\n", a, b);
return(0);
}
这到底是怎么一回事？

Answer 1

“引用传递做函数参数”是C++的特性，C语言不支持。

// C语言要这样写：
void Exchg3(Rect *x, Rect *y){ /* ... */ }

// 然后传递指针：
Exchg3(&a, &b);

Answer 2

不保存.cpp 编译的时候就用c++ 编译器
你用.c编译的时候用c编译器

还有就是c 不支持引用

Answer 3

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
typedef struct { int x, y, width, height; } Rect;
void Exchg3(Rect x, Rect y)
{
int tmp = x.height;
x.height = y.height;
y.height = tmp;
printf("x.height = %d,y.height = %d\n", x.height, y.height);
}
int main()
{
Rect a = { 1, 2, 3, 4 };
Rect b = { 5, 6, 7, 8 };
Exchg3(a, b);
printf("a = %d, b = %d\n", a, b);
return(0);
}
C里面这样写