#include <stdio.h>
#include <stdlib.h>
int swap1();
int swap2();
int main()
{
int m = 0, n = 1;
printf("Starting values: m, n = %d, %d\n", m, n);
swap1(m, n); /* call by value */
printf("After swap1(): m, n = %d, %d\n", m, n);
swap2(&m, &n); /* call by reference */
printf("After swap2(): m, n = %d, %d\n\n", m, n);
return 0;
}
/* swap1: swaps copies of values */
int swap1(int x, int y)
{
int temp;
temp = x;
x = y;
y = temp;
return 0;
}
/* swap2: swaps values at the addresses given */
int swap2(int *x, int *y)
{
int temp;
temp = *x;
*x = *y;
*y = temp;
return 0;
为什么两个函数输出一个是01，一个是10啊？

#include <stdio.h>
#include <stdlib.h>
int swap1();
int swap2();
int main()
{
int m = 0, n = 1;
printf("Starting values: m, n = %d, %d\n", m, n);
swap1(m, n); /* call by value */
printf("After swap1(): m, n = %d, %d\n", m, n);
swap2(&m, &n); /* call by reference */
printf("After swap2(): m, n = %d, %d\n\n", m, n);
return 0;
}
/* swap1: swaps copies of values */
int swap1(int x, int y)//获得m n的值，此处x y相当于m n的副本，对xy的任何改动不影响原来的mn
{
int temp;
temp = x;
x = y;
y = temp;
return 0;
}
/* swap2: swaps values at the addresses given */
int swap2(int *x, int *y)//获得m n 的地址，通过地址直接操纵mn中的数，故能交换mn
{
int temp;
temp = *x;
*x = *y;
*y = temp;
return 0;

是不是比如return x；然后x的值就对m产生影响了啊？

着就是函数参数的值传递，和地址传递的区别。

不懂你啥意思……

没有影响