当输入太长时，如何限制C中的scanf函数打印错误? [英] How to limit scanf function in C to print error when input is too long?

问题描述

我想限制 scanf 函数，所以当我输入例如一个 char* 数组 <String...> 超过 30 个字符时，它不会得到它和我的输出会出错.

I want to limit the scanf function so when I enter for example a char* array <String...> that has more then 30 characters, it will not get it and my output will be error.

我得到了使用 [^n] 或类似内容的提示，但我不明白该怎么做?我知道我可以使用 scanf("%30s"..) 但我不希望输入有效而只是错误.

I got a hint to use [^n] or something like that but I don't understand how to do it? I know that I can use scanf("%30s"..) but I don't want the input to be valid and just the error.

任何帮助都会很棒.

推荐答案

如果你必须使用 scanf 那么我相信你能做的最好的事情就是使用宽度说明符，例如:"%31s"，正如你已经提到的，然后使用strlen来检查输入的长度，如果输入的长度超过你的长度，则丢弃该字符串并报告错误限制.

If you must use scanf then I believe that the best that you can do is use the width specifier with something like: "%31s", as you've already mentioned, then use strlen to check the length of the input, and discard the string and report an error if the input is longer than your limit.

或者可能通过在格式字符串中额外使用 %n 来跳过 strlen，例如"%31s%n".

Or possibly skip the strlen by additionally using an %n in your format string, e.g. "%31s%n".

使用类似 %[^ ] 代替 %s 的格式字符串只是指示函数继续读取直到换行，同时消耗其他空白字符道路.如果您希望允许输入包含空白字符，这将非常有用.

A format string using something like %[^ ] in place of %s simply instructs the function to continue reading until a newline, consuming other whitespace characters along the way. This is useful if you want to allow the input to include whitespace characters.

查看 scanf 的文档(这里是手册页的副本).

Review the docs for scanf (here's a copy of the man page).

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