getChar（）跳过c程序中的输入

Question

\n

（enter）仍然保留在缓冲区中，然后在while循环中以第一个

getchar()

读，这在缓冲区中没有任何东西要在第二个

getchar()

中读取，但仍然以某种方式进入循环，“嗯，那是6吗？”执行，我不明白如何。我将代码修复了我写的目的，但我仍然想知道是什么原因导致了这个确切的问题。

编码：

#include <stdio.h>

int main() {
    int guess = 1;
    char c;
    
    printf("The program guesses the number between 1 and 100.\n");
    printf("Respond it with Y if the guess is correct and N if the guess is wrong.\n");
    printf("is your number %d?\n", guess);
    
    while((c = getchar()) != 'Y' || (c = getchar()) != 'y') {
        printf("well, is it %d then?\n", ++guess);
        if((c = getchar()) == '\n')
            continue;
    }
    printf("I knew it was %d", guess);

    return 0;
}

输出：

The program guesses the number between 1 and 100.
Respond with Y if the guess is correct and N if the guess is wrong.
Is your number 1?
n
Well, is it 2 then?
n
Well, is it 3 then?
n
Well, is it 4 then?
nn
Well, is it 5 then?
Well, is it 6 then?
n
Well, is it 7 then?

I期望WARE循环终止，提示最后一份

printf()

最终导致该程序结束，而是随着此输出而结束。

The program guesses the number between 1 and 100.
Respond it with Y if the guess is correct and N if the guess is wrong.
is your number 1?
n
well, is it 2 then?
n
well, is it 3 then?
n
well, is it 4 then?
nn
well, is it 5 then?

您有一个无尽的循环，因为所有字母并非“ y”或不是“ y”。
尝试寻找不是“ y”而不是“ y”的字母。

如果您改变了您选择打电话两次的问题，例如您必须两次输入“ N”才能获得任何反馈。

Answer 1

我建议在循环结束时进行阅读和检查。 to做到这一点，我建议从

getchar()
切换到

while

。