如何在 Linux 中使用 crypt() 方法？
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how to use crypt( ) method in Linux?
提问by user1198331
I just want to use crypt()to generate an encrypted password,and I write a demo which invoke the crypt()method.
Here is my code
我只是想用来crypt()生成一个加密的密码，我写了一个调用该crypt()方法的演示。这是我的代码
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
int main()
{
printf("%s\n",crypt("abc","ab"));
exit(0);
}
I compile it using "gcc tem.c -lcrypt'and when I run it, everything seems right, but a "segment error" shows up. so please tell me what's wrong with this simple program?
我使用编译它"gcc tem.c -lcrypt'，当我运行它时，一切似乎都正确，但出现了“段错误”。所以请告诉我这个简单的程序有什么问题？
采纳答案by Some programmer dude
If you compile with the flag -Wallyou will see why.
如果您使用该标志进行编译，-Wall您就会明白为什么。
If you read the manual pageyou will see that it uses #define _XOPEN_SOURCEbefore including <unistd.h>. It should actually be defined before including anyheader.
如果您阅读手册页，您将看到它#define _XOPEN_SOURCE在包含<unistd.h>. 它实际上应该在包含任何标题之前定义。
If you don't define _XOPEN_SOURCEthen the cryptfunction will not be prototyped. Then the compiler doesn't know what the actual return type is, or the types and number of arguments. So it will assume that the function returns an intand your printfexpects a string, so there will be a type mismatch that causes the crash.
如果您不定义，_XOPEN_SOURCE则该crypt函数将不会被原型化。然后编译器不知道实际的返回类型是什么，或者参数的类型和数量。所以它会假设函数返回一个int并且你printf期望一个字符串，所以会有一个导致崩溃的类型不匹配。
回答by teppic
You need this:
你需要这个：
#define _XOPEN_SOURCE
at the top of your source file, before any #include.
在源文件的顶部，在任何#include.
Alternatively compile with the gcc option -D_XOPEN_SOURCE.
或者使用 gcc 选项编译-D_XOPEN_SOURCE。
回答by Mark Stanislav
Looks like it could be related to crypto library support.
看起来它可能与加密库支持有关。
Try adding:
尝试添加：
#include <crypt.h>
[mstanislav@pardalislabs ~]$ gcc tem.c -lcrypt
[mstanislav@pardalislabs ~]$ ./a.out
abFZSxKKdq5s6
Looks good for me!
对我来说看起来不错！
