密码函数和链接错误“未定义对“密码"的引用"； [英] crypt function and link error "undefined reference to 'crypt'"

问题描述

我在c语言中使用了crypt函数来加密给定的字符串. 我已经编写了以下代码，

I have used the crypt function in c to encrypt the given string. I have written the following code,

#include<stdio.h>
#include<unistd.h>

int main()
{
    printf("%s\n",crypt("passwd",1000));
}

但是上面的代码抛出了一个错误，未定义对"crypt"的引用". 上面的代码有什么问题.

But the above code threw an error ,"undefined reference to `crypt'". What is the problem in the above code.

提前谢谢.

推荐答案

如果要使用crypt()函数，则需要链接到提供它的库.将-lcrypt添加到您的编译命令中.

If you want to use the crypt() function, you need to link to the library that supplies it. Add -lcrypt to your compile command.

较早版本的glibc为此提供了一个libcrypt库，并在<unistd.h>中声明了该功能-为了对此支持进行编译，您可能还需要在代码中定义_XOPEN_SOURCE或_GNU_SOURCE包括<unistd.h>.

Older versions of glibc supplied a libcrypt library for this purpose, and declared the function in <unistd.h> - to compile against this support, you may also need to define either _XOPEN_SOURCE or _GNU_SOURCE in your code before including <unistd.h>.

较新版本的glibc不提供libcrypt-而是由单独的libxcrypt提供.您仍然与-lcrypt链接，但是该函数改为在<crypt.h>中声明.

Newer versions of glibc don't supply libcrypt - it is instead provided by a separate libxcrypt. You still link with -lcrypt, but the function is instead declared in <crypt.h>.

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