c - Kochan Book : Programming in C i + j - i % j 中的一项练习遇到问题

Question

这是我的 C 源代码:

details that stackoverflow want to see or i can't edit

#include <stdio.h>

int main (void) {

    int  i, j;

    int Next_multiple = (i + j) - (i % j);

    i = 365;
    j = 7;

    printf("Solution for i == 365 j == 7 = %i \n", Next_multiple);

    i = 12258;
    j = 23;

    printf("Solution for i == 12258 j == 23 = %i \n", Next_multiple);

    i = 996;
    j = 4;

    printf("Solution for i== 996 j == 4 = %i \n", Next_multiple);

    return 0;
}

这就是输出:

.exe(Visual Studio 工具中的 Windows cmd)

Solution for i == 365 j == 7 = 18039659

Solution for i == 12258 j == 23 = 18039659

Solution for i== 996 j == 4 = 18039659

Answer 1

1. 当您使用未初始化的局部变量 i 和 j 时，您会出现未定义的行为。

然后显示 3 倍相同的计算变量(使用这些随机值)。该如何整理呢？

1 将此函数放在 main 函数之前。

  int Next_multiple(int i, int j)
  {
     return ( i + j) - (i % j);
   }

改变

printf("Solution for i == 12258 j == 23 = %i \n", Next_multiple);

至

  printf("Solution for i == 12258 j == 23 = %i \n", Next_multiple(i, j));

并删除

 int Next_multiple = (i + j) - (i % j);

<小时/>

#include <stdio.h>

int Next_multiple(int i, int j)
{
   return ( i + j) - (i % j);
}
int main () {

    int  i, j;

    i = 365;
    j = 7;

    printf("Solution for i == 365 j == 7 = %i \n", Next_multiple(i,j));

    i = 12258;
    j = 23;

    printf("Solution for i == 12258 j == 23 = %i \n", Next_multiple(i,j));

    i = 996;
    j = 4;

    printf("Solution for i== 996 j == 4 = %i \n", Next_multiple(i,j));

    return 0;
}