为什么这个程序起不到互换的作用呢
#include <stdio.h>#include <stdlib.h>
return swap(int a,int b)
{
intt;
t = a;
a = b;
b = t;
}
int main()
{
int a = 3;
int b = 5;
swap(a,b);
printf("a = %d,b = %d",a,b);
return 0;
}
本帖最后由 傻眼貓咪 于 2022-1-19 11:34 编辑
参数用指针
注:return 在函数里面,不是外面,这里不需要用 return #include <stdio.h>
void swap(int *a, int *b){
int temp = *a;
*a = *b;
*b = temp;
}
int main(){
int a = 13, b = 7;
swap(&a, &b);
printf("a: %d\n", a);
printf("b: %d\n", b);
return 0;
} #include <stdio.h>
void swap(int *a,int *b)
{
intt;
t = *a;
*a = *b;
*b = t;
}
int main()
{
int a = 3;
int b = 5;
swap(&a,&b);
printf("a = %d,b = %d",a,b);
return 0;
} 需使用指针
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