在Unix中使用sbrk系统调用

Question

What is the use of sbrk system call in Unix and how is it to be used in C? Most of them say that malloc uses the sbrk system call to allocate memory. So, how does the sbrk system call allocate memory? What are the arguments and return type of sbrk?

Answer 1

sbrk() takes a positive integer as an argument. It increases the size of the current process's heap by that number of bytes, then returns a void * to the new end of the heap.

However, the information you're reading is outdated . To quote the Mac OS X manual page for this system call:

The brk and sbrk functions are historical curiosities left over from earlier days before the advent of virtual memory management.

Modern systems do not use sbrk() for memory management, as it is incredibly limited. In particular, it cannot manage a heap which is not contiguous in memory! Instead, they use the mmap() system call to map new ranges of memory as needed.

Answer 2

Function definition: void *sbrk(intptr_t increment);

sbrk() increments the program's data space by increment bytes. Calling sbrk() with an increment of 0 can be used to find the current location of the program break.

From http://linux.die.net/man/2/sbrk