由malloc(0)返回的跟随指针 [英] Following pointer returned by malloc(0)

问题描述

我正在尝试理解部分代码.为了简化说明并避免不必要的混乱，我遗漏了很多代码.

I am trying to understand a portion of code. I am leaving out a lot of the code in order to make it simpler to explain, and to avoid unnecessary confusion.

typedef void *UP_T;

void FunctionC(void *pvD, int Offset) {
    unsigned long long int temp;
    void *pvFD = NULL;

    pvFD = pvD + Offset;
    temp = (unsigned long long int)*(int *)pvFD;
}

void FunctionB(UP_T s) {
    FunctionC(s, 8);
}

void FunctionA() {
    char *tempstorage=(char *)malloc(0);
    FunctionB(tempstorage);
}

int main () {
    FunctionA();
    return 0;
}

就像我说的那样，我遗漏了大量代码，因此出现的功能无用，因为它们只有两行代码.

Like I said, I am leaving out a ton of code, hence the functions that appear useless because they only have two lines of code.

什么是temp?那就是让我困惑的地方.当我运行类似于此代码的代码，并在整个过程中使用printf()语句时，会得到pvD的随机数，而pvFD是该随机数加8.

What is temp? That is what is confusing me. When I run something similar to this code, and use printf() statements along the way, I get a random number for pvD, and pvFD is that random number plus eight.

但是，我也可能打印了错误的值(使用%llu而不是%d或类似的方法).我很确定这是指向tempstorage加8的内存位置的指针.这是正确的吗?我只是想确定一下，然后再继续执行该假设.

But, I could also be printing the values incorrectly (using %llu instead of %d, or something like that). I am pretty sure it's a pointer to the location in memory of tempstorage plus 8. Is this correct? I just want to be certain before I continue under that assumption.

推荐答案

标准指定malloc(0)返回NULL或有效的指针，但永远不要取消引用该指针.对实际的实现没有任何限制，因此您不能依赖于返回的指针是另一个加8.

The standard specifies that malloc(0) returns either NULL or a valid pointer, but that pointer is never to be dereferenced. There aren't any constraints regarding the actual implementation, so you can't rely on the returned pointer being another plus 8.

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