[英]Why procedure if in C doesn't work with char
I am writing a simple quiz in C (using CodeBlocks 13.12)我正在用 C 编写一个简单的测验(使用 CodeBlocks 13.12)
It compiles, but doesn't work in second question.它编译,但在第二个问题中不起作用。 Whatever I will input, it always give answer ' that's sad '.
无论我要输入什么,它总是给出'那很悲伤'的答案。 I can't understand what is wrong.
我不明白出了什么问题。 I came to this, where if I comment line 13 ( scanf("%d", &age); ) it's starting works ok for second question.
我来到了这里,如果我评论第 13 行( scanf("%d", &age); ),它开始可以用于第二个问题。 What the problem is?
问题是什么?
#include <iostream>
#include <stdio.h>
#include <windows.h>
#include <clocale>
int main()
{
int age;
char S1;
printf("How old is your dog? \n");
scanf("%d", &age);
if (age <= 7)
{
printf(" very young. the end \n");
return 0;
}
else
{
printf("old dog. \n \n");
}
//question2
printf("Do you like dogs? y/n \n");
scanf("%c%c", &S1);
if (S1 == 'y')
{
printf("hey, that's nice \n");
}
else
{
printf(" that's sad :( . \n");
return 0;
}
return 0;
}
You cause undefined behavior by你导致未定义的行为
scanf("%c%c", &S1);
scanf
reads two char
s, one stored in S1
, one stored in some location on the stack because scanf
expects a second char*
to be supplied. scanf
读取两个char
,一个存储在S1
,一个存储在堆栈上的某个位置,因为scanf
期望提供第二个char*
。
If your intention is to ignore the newline following the actual character, write如果您打算忽略实际字符后面的换行符,请写
scanf("%c%*c", &S1);
Change the second scanf()
to将第二个
scanf()
更改为
scanf(" %c", &S1);
This would escape the left out newline character \\n
in the input buffer.这将转义输入缓冲区中遗漏的换行符
\\n
。
Plus, you are reading one char
in this.另外,您正在阅读其中的一个
char
。 So you need only one %c
所以你只需要一个
%c
scanf("%c", &S1);
是输入一个字符的正确方法,
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