c - scanf 不等待输入就返回 0

Question

我从来没有用 C 编程，今天我必须编写一些小代码。程序非常简单 - 我想将两个整数相加。但是当我试图检查给定的输入是否是数字并且第一个 scanf 返回 0 时，第二个 scanf 也返回 0 而无需等待输入。代码:

int main()
{
    int a = 0;
    int b = 0;
    printf("Number a:\n");
    if (scanf("%d", &a) != 1)
    {
        printf("Not a number. a=0!\n");
        a = 0;
    }
    printf("Number b:\n");
    if (scanf("%d", &b) != 1)
    {
        printf("Not a number. b=0!\n");
        b = 0;
    }
    printf("%d\n", a+b);
    return 0;
}

Answer 1

第一个 fscanf() 未能转换为数字的输入仍在标准输入缓冲区中等待，并导致第二个 fscanf() 失败。尝试丢弃有问题的输入并重新提示用户:

#include <stdio.h>

int main(void) {
    int a = 0;
    int b = 0;
    int c;
    printf("Number a:\n");
    while (scanf("%d", &a) != 1) {
        printf("Not a number, try again:\n");
        while ((c = getchar()) != EOF && c != '\n')
            continue;
        if (c == EOF)
            exit(1);
    }
    printf("Number b:\n");
    while (scanf("%d", &b) != 1) {
        printf("Not a number, try again:\n");
        while ((c = getchar()) != EOF && c != '\n')
            continue;
        if (c == EOF)
            exit(1);
    }
    printf("%d\n", a + b);
    return 0;
}

使用实用函数分解代码使其更加清晰:

#include <stdio.h>

int get_number(const char *prompt, int *valp) {
    printf("%s:\n", prompt);
    while (scanf("%d", valp) != 1) {
        printf("Not a number, try again:\n");
        while ((c = getchar()) != EOF && c != '\n')
            continue;
        if (c == EOF)
            return 0;
    }
    return 1;
}

int main(void) {
    int a, b;

    if (!get_number("Number a", &a) || !get_number("Number b", &b)) {
         return 1;
    }
    printf("%d\n", a + b);
    return 0;
}