打印空指针的位 [英] print bits of a void pointer

问题描述

如果我创建一个void指针，并为该void指针分配了一段内存，那么如何打印出刚分配的各个位?

If I create a void pointer, and malloc a section of memory to that void pointer, how can I print out the individual bits that I just allocated?

例如:

void * p;
p = malloc(24);
printf("0x%x\n", (int *)p);

我希望上面显示我刚刚分配的24位.

I would like the above to print the 24 bits that I just allocated.

推荐答案

size_t size = 24;
void *p = malloc(size);

for (int i = 0; i < size; i++) {
  printf("%02x", ((unsigned char *) p) [i]);
}

当然，它会调用未定义的行为(malloc分配的对象的值具有不确定的值).

Of course it invokes undefined behavior (the value of an object allocated by malloc has an indeterminate value).

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