(void *)-1的含义 [英] significance of (void*) -1

问题描述

我正在查看sbrk系统调用的文档，发现此:

I was looking at the documentation of sbrk system call and found this:

成功后，sbrk()返回上一个程序中断. (如果增加了中断，则此值是指向新分配的内存的起点的指针).出现错误时，将返回(void *) -1，并将errno设置为ENOMEM.

On success, sbrk() returns the previous program break. (If the break was increased, then this value is a pointer to the start of the newly allocated memory). On error, (void *) -1 is returned, and errno is set to ENOMEM.

现在

1. (void *) -1的意义是什么?

它指向的确切内存地址是什么? (如果有的话)

What is the exact memory address it points to? (if it does at all)

如何保证(void *) -1不是有效地址 可以由sbrk()成功返回吗?

How is it guaranteed that (void *) -1 is not a valid address that can be returned by sbrk() on success?

推荐答案

(void *) -1 == (size_t) -1

在32位计算机上为0xFFFFFFFF，在64位计算机上为0xFFFFFFFFFFFFFFFF，该无效地址应该大于任何其他地址.

It's 0xFFFFFFFF on 32 bit machine and 0xFFFFFFFFFFFFFFFF on 64 bit machine, an invalid address that is supposed to be bigger than any other address.

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