用scanf()读取的字符串大小不超过 [英] Read no more than size of string with scanf()

问题描述

对于我的班级，我必须使用 scanf .因此，推荐其他输入方式不是我正在寻找的解决方案(如果有涉及scanf的解决方案).

for my class I have to use scanf. So recommending other ways of input is not the solution I am looking for (if there is one that involves scanf).

如果我正在阅读用户输入的小项目(例如游戏).可以说我问您想玩吗?这将接受 yes 或 no 答案.所以我写了一些像这样的简单代码:

If I am reading in user input for a small project (for example, a game). Lets say I ask would you like to play? This would accept a yes or no answer. So i write up some simple code like this:

#include <stdio.h>

int main(void)
{
     char string[3]; //The max number of letters for "yes".

     printf("Would you like to play?");
     scanf("%s", string);
}

因此，此代码应仅要求他们输入是或否.我将我的 char数组的长度设置为大小 3 .这样，它足以容纳 yes 和 no .但是，如果有人输入诸如 yesss 之类的无效输入，我知道以后如何比较字符串以处理此类事件，但是从技术上/可能不会覆盖我声明的其他局部变量，因为它会超出我的数组长度吗?如果是这样，是否有一种方法可以限制3个输入字符或其他字符呢?如果不是，为什么/怎么知道只输入3的大小?

So this code should simply ask them to input yes or no. I am setting the length of my char array to size 3. This way it is large enough to hold yes and also no. But if someone were to enter invalid input such as yesss, I know how to compare the string afterwards to handle such an event, but wouldn't this technically/possibly overwrite other local variables I have declared because it would extend outside the length of my array? If so, is there a way to handle this to restrict 3 input characters or something? And if not, why/how does it know to only input for the size of 3?

推荐答案

您的数组必须能够容纳四个 char ，因为它还必须包含0终止符.固定后，以格式指定最大长度，

Your array needs to be able to hold four chars, since it must also contain the 0-terminator. With that fixed, specifying a maximal length in the format,

scanf("%3s", string);

确保 scanf 读取的字符数不超过3个.

ensures that scanf reads no more than 3 characters.

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