gcc在Debian中的奇怪行为

char * stft (const char *fmt, ...) {

    va_list items;
    char *out;
    int magic = 0; // <-- here magic?

    va_start (items, fmt);
    vsprintf (out, fmt, items);
    va_end (items);

    return out;

}

Use like:

使用如下：

char *str = stft ("%s-%s %s", a, b, c);

This is working solution? if delete unused "magic" variable - I have Segmentation fault after return string. What doing wrong?

这是有效的解决方案？如果删除未使用的“魔术”变量 - 返回字符串后我有分段错误。做错了什么？

$ gcc --version gcc (Debian 4.4.5-8) 4.4.5

$ gcc --version gcc（Debian 4.4.5-8）4.4.5

$ uname -a Linux deep-station (squeeze) 2.6.32-5-686 #1 SMP Fri May 10 08:33:48 UTC 2013 i686 GNU/Linux

$ uname -a Linux深站（挤压）2.6.32-5-686＃1 SMP Fri 5月10日08:33:48 UTC 2013 i686 GNU / Linux

1 个解决方案

#1

You are trying to write to an uninitialized pointer out. That's why you crash. It is badly undefined behaviour. The magic is coincidental; it does not make the behaviour any better defined.

您正在尝试写入未初始化的指针。这就是你崩溃的原因。这是严重未定义的行为。巧妙是巧合;它不会使行为更好地定义。

It is probably best to use vsnprintf():

最好使用vsnprintf（）：

char *out = malloc(256);
...
vsnprintf(out, 256, fmt, items);
...
return out;

Or something similar.

或类似的东西。

You can improve this. For example:

你可以改善这一点。例如：

char *stft(const char *fmt, ...)
{
    va_list items;

    va_start(items, fmt);
    int length = vsnprintf(0, 0, fmt, items);
    va_end(items);
    char *out = malloc(length+1);
    if (out != 0)
    {
        va_start(items, fmt);
        vsnprintf(out, length+1, fmt, items);
        va_end(items);
    }

    return out;
}

Make sure you free the allocated memory in the calling code.

确保在调用代码中释放已分配的内存。

#1