Question

我正在尝试运行一个简单的代码，它从用户那里获取两个数字并打印它们的总和那是：

#include <stdio.h>
#include <conio.h>

void main() {
    int n1 = 0, n2 = 0, sum = 0;
    printf("Insert two numbers:\n");
    scanf("%d %d", n1, n2);
    sum = n1 + n2;
    printf("The sum of %d and %d is - %d\n", n1, n2, sum);
    printf("%d + %d = %d", n1, n2, sum);
    getch();
}

我正在尝试运行它，当我执行第一个打印命令时，当我按Enter键进行扫描时崩溃，我正在使用code :: blocks 13.12来编译和运行谢谢！：）

Answer 1

您的程序崩溃是因为您正在将数字扫描到错误的内存地址。改变

scanf("%d %d", n1, n2);

到

scanf("%d %d", &n1, &n2);

Answer 2

您的程序无法运行，因为scanf为int个"%d"指针设置n1，因此请传递n2和{{1}的地址而不是n1和n2。

#include <stdio.h>
#include <conio.h>

void main() {
    int n1 = 0, n2 = 0, sum = 0;
    printf("Insert two numbers:\n");
    scanf("%d %d", &n1, &n2);
    /*             ^    ^ address of operater, take the addreess of n2 and pass it */
    /*             | address of operater, take the addreess of n1 and pass it */
    sum = n1 + n2;
    printf("The sum of %d and %d is - %d\n", n1, n2, sum);
    printf("%d + %d = %d", n1, n2, sum);
    getch();
}

但不仅如此，您应该检查实际上读取的数字，否则您的程序将调用未定义的行为，以确保您成功读取整数，您应该检查它返回的scanf的返回值匹配的参数数量，所以

#include <stdio.h>
#include <conio.h>

void main() {
    int n1 = 0, n2 = 0, sum = 0;
    printf("Insert two numbers:\n");

    /* if it equals two it succeeded, since we requested two integers */
    if (scanf("%d %d", &n1, &n2) == 2)
    {
        sum = n1 + n2;
        printf("The sum of %d and %d is - %d\n", n1, n2, sum);
        printf("%d + %d = %d", n1, n2, sum);
    }
    else
    {
        printf("error reading 2 integers\n");
        return 1; /* non zero means error */
    }
    getch();
    /* return success from main() */
    return 0;
}

无法运行基本的C程序

2 个答案: