UC知道C语言构造体的调用函数程序，请高手修改，真诚请你贴上能成功运行的程序
来源：百度知道 编辑：UC知道 时间：2024/07/04 07:02:16
我想通过构造体函数的形式来表示复数，然后通过调用函数，输出函数的实部与虚部，但是还没有运行成功，敬请赐教。特别说明，我用的是microsoft visual c++ 2008.
#include<stdio.h>
struct well{
int re;
int im;
};
typedef struct well well;
int add(well *a, well *b, well *c);
int main(void)
{
int a, b, c;
add(a, b, c);
printf("a->re = %d\na->im = %d\n", a->re, a->im);
}
int add(well *a, well *b, well *c)
{
a->re = b->re + c->re;
a->im = b->im + c->im;
return a->re, a->im;
}
我还忘记初始化了。
做一下初始化。
（*a = 0 + 0r;
*b = 2 + 3r;
*c = 1 + 4r;
这里r表示虚部）
that is
初始化
a->re = 0;
a->im = 0;
b->re = 2;
b->im = 3;
c->re = 1;
c->im = 4;
#include<stdio.h>
struct well{
int re;
int im;
};
typedef struct well well;
int add(well *a, well *b, well *c);
int main(void)
{
int a, b, c;
add(a, b, c);
printf("a->re = %d\na->im = %d\n", a->re, a->im);
}
int add(well *a, well *b, well *c)
{
a->re = b->re + c->re;
a->im = b->im + c->im;
return a->re, a->im;
}
我还忘记初始化了。
做一下初始化。
（*a = 0 + 0r;
*b = 2 + 3r;
*c = 1 + 4r;
这里r表示虚部）
that is
初始化
a->re = 0;
a->im = 0;
b->re = 2;
b->im = 3;
c->re = 1;
c->im = 4;
根据你对add函数的定义，main函数的int a, b,c ; add(a,b,c);这两行是不对的。另外你没有对b和c初始化， 这样的计算也没有意义，因为这样的自动变量值是随机的。
可以改成以下这个样子。
#include<stdio.h>
struct well{
int re;
int im;
};
typedef struct well well;
int add(well *a, well *b, well *c);
int main(void)
{
well a, b, c;
b.re = 3; b.im = 4;
c.re = 4; c.im = 7;
add(&a, &b, &c);
printf("a->re = %d\na->im = %d\n", a.re, a.im);
}
int add(well *a, well *b, well *c)
{
a->re = b->re + c->re;
a->im = b->im + c->im;
return a->re, a->im;
}
#include<stdio.h>
struct rwell{
int re;
int im;
}well;
well owell;
well *add(well a, well b);
int main(void)
{
well a, b, *c;
scanf("a.re = %d\na.im = %d\n", a.re, a.im);
scanf("b.re = %d\nb.im = %d\n&quo