C: char 的 scanf 没有按预期工作 [英] C: scanf for char not working as expected

问题描述

我最近在我的电脑上运行了一个 c 程序.它有一个 for 循环，其中扫描了一些 char d.for 循环运行 3 次.在每次运行期间，它会打印运行计数，然后扫描 char d 的值.程序如下

I was recently running a c program in my PC. It have a for loop in which some char d is scanned. The for loop runs for 3 times. During each running it prints the the count of running and then scans the value of char d. The program is as follows

#include<stdio.h>

int main(){
    int f;
    char d;
    for(f=0;f<3;f++){
        printf("Choice %d
", f);
        scanf("%c", &d);
    }
    return 0;
}

现在的问题是，当我运行程序时，当 f 为 1 时，for 会跳过 scanf 部分.现在，如果我将代码更改如下

Now the trouble is that when I run the program, the for skips the scanf part when f is 1. Now if i changed the code as follows

#include<stdio.h>

int main(){
    int f;
    int d;
    for(f=0;f<3;f++){
        printf("Choice %d
", f);
        scanf("%d", &d);
    }
    return 0;
}

现在程序运行良好.每一次for循环迭代都会执行scanf.

Now the program works fine. and scanf is executed for every iteration of for loop.

这里似乎有什么问题?我的意思是当 d 是 int 类型时它可以正常工作，但是当 d 是 char 类型时它不能正常工作.

What does seem to be the problem here? I mean when d is of type int it works fine, but when d is of type char it does not work correctly.

推荐答案

你必须改变

scanf("%c", &d);

到

scanf(" %c", &d);
       ^
       |

否则，scanf()会考虑之前输入的ENTER键按下.

Otherwise, scanf() will consider the previously entered ENTER key press.

注意:

1. ENTER 键按下生成一个 ，这是%c 格式说明符的有效输入.在 %c 之前添加一个空格告诉 scanf() 忽略所有前导的类似空格的输入(包括之前存储的 )并读取stdin 中的第一个非空白字符.

1. ENTER key press generates a , which is a vaild input for %c format specifier. Adding a space before %c tells scanf() to ignore all leading whitespace-like inputs (including that previously stored ) and read the first non-whitespace character from stdin.

对于 %d 格式说明符的情况，它会在扫描数字输入之前消耗(并忽略)任何前导的类似空格的输入，因此第二种情况不会遇到任何问题.

As for the case with %d format specifier, it consumes (and ignores) any leading whitespace-like inputs before scanning for numeric inputs, so the second case does not suffer any issues.

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