程序不会在 scanf("%c", &ch) 行停止，为什么? [英] The program doesn't stop on scanf("%c", &ch) line, why?

问题描述

程序不会在 scanf("%c", &ch) 行停止.为什么会这样有人可以向我解释一下

the program doesnt stop on scanf("%c", &ch) line. why does it happens sombody can please explain this to me

#include<stdlib.h>
#include<stdio.h>

struct list {
   char val;
   struct list * next;
};

typedef struct list item;

void main()
{
    char ch;
    int num;

    printf("Enter [1] if you want to use linked list or [2] for realloc
");  
    scanf("%d", &num);
    if(num == 2)
    {
        scanf("%c", &ch); 
        printf("%c", ch);
    }
}

推荐答案

假设您在阅读 num 时输入 2.实际输入流将为 2 ( 是换行符).2 进入 num，还有 进入 ch.为避免这种情况，请在格式说明符中添加一个空格.

Let's say you input 2 when you're reading for num. The actual input stream will be 2 ( is the newline character). 2 goes into the num, and there remains , which goes into ch. To avoid this, add a whitespace in format specifier.

scanf(" %c", &ch); 

这将忽略任何空格、换行符或制表符.

This will ignore any whitespaces, newlines or tabs.

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