scanf不超过缓冲区溢出 [英] scanf not to exceed buffer overrun

问题描述

我有一个缓冲区，我不希望用户输入的字符超出缓冲区可以容纳的字符数（以避免缓冲区溢出）。

I have a buffer and I don't want the user to enter more characters than the buffer can hold (to avoid a buffer overrun).

我正在使用 scanf 并执行以下操作：

I am using scanf and have done like this:

char buffer[30] = {'\0'};
scanf("%30s", buffer);

但是，如果用户输入的字符数超过30，我知道我会受到保护。但是，如果用户输入超过30，缓冲区将以null终止吗？

However, I know I am protected if the user enters more than 30. However, if the user enters more than 30, will the buffer be null terminated?

推荐答案

scanf（）带有％s转换说明符的缓冲区会添加一个终止的空字符。

scanf() with a "%s" conversion specifier adds a terminating null character to the buffer.

但是，您要输入30个字符，实际上表示31，只能容纳30个空间。您应该使用最大字段宽度为29。

But, you're asking for 30 characters, which really means 31 and only have space for 30. You should use a maximum field width of 29.

char buffer[30] = {'\0'};
scanf("%29s", buffer);

还要注意，转换说明符％c 的工作原理与％s 相似，但是不会添加终止的空字符，也不会从输入中丢弃空格。取决于您的期望，它可能比使用％s更好。

Also note that the conversion specifier "%c" works pretty much like "%s", but does not add the terminating null character and does not discard space from the input. Depending on what you expect, it might be better than using "%s".

char buffer[30] = {'\0'};
scanf("%29c", buffer);
buffer[29] = '\0';

这篇关于scanf不超过缓冲区溢出的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！