c - 在 C 中生成随机 UUID

Question

我将如何在 C 中生成基于熵的 UUID 并将其存储为字符串(字符指针)？

我希望有一种简单的方法可以在内部执行此操作，但如果没有，system("uuidgen -r") 也可以。

Answer 1

此功能由 libuuid 提供. (Debian 上的软件包 libuuid1 和 uuid-dev。)

这是一个生成基于熵的(随机)UUID 并将其写入 stdout 的简单程序, 然后以状态 0 退出.

/* For malloc() */
#include <stdlib.h>
/* For puts()/printf() */
#include <stdio.h>
/* For uuid_generate() and uuid_unparse() */
#include <uuid/uuid.h>


/* Uncomment to always generate capital UUIDs. */
//#define capitaluuid true

/* Uncomment to always generate lower-case UUIDs. */
//#define lowercaseuuid true

/*
 * Don't uncomment either if you don't care (the case of the letters
 * in the 'unparsed' UUID will depend on your system's locale).
 */


int main(void) {
    uuid_t binuuid;
    /*
     * Generate a UUID. We're not done yet, though,
     * for the UUID generated is in binary format 
     * (hence the variable name). We must 'unparse' 
     * binuuid to get a usable 36-character string.
     */
    uuid_generate_random(binuuid);

    /*
     * uuid_unparse() doesn't allocate memory for itself, so do that with
     * malloc(). 37 is the length of a UUID (36 characters), plus '\0'.
     */
    char *uuid = malloc(37);

#ifdef capitaluuid
    /* Produces a UUID string at uuid consisting of capital letters. */
    uuid_unparse_upper(binuuid, uuid);
#elif lowercaseuuid
    /* Produces a UUID string at uuid consisting of lower-case letters. */
    uuid_unparse_lower(binuuid, uuid);
#else
    /*
     * Produces a UUID string at uuid consisting of letters
     * whose case depends on the system's locale.
     */
    uuid_unparse(binuuid, uuid);
#endif

    // Equivalent of printf("%s\n", uuid); - just my personal preference
    puts(uuid);

    return 0;
}

uuid_unparse()不分配它自己的内存；为避免执行时出现段错误，您必须使用 uuid = malloc(37); 手动执行此操作(您还可以将 UUID 存储在该长度的字符数组中:char uuid[37];)。确保使用 -luuid 进行编译这样链接器就知道 uuid_generate_random()和 uuid_unparse()在 libuuid 中定义.