为什么函数 brk() 的参数是 void* 而不是 int 类型？

Question

我正在查看 Linux 指南中函数 int brk() 的文档：

SYNOPSIS
   #include <unistd.h>

   int brk(void *addr);

   void *sbrk(intptr_t increment);



Feature Test Macro Requirements for glibc (see feature_test_macros(7)):

   brk(), sbrk():
       Since glibc 2.19:
           _DEFAULT_SOURCE ||
               (_XOPEN_SOURCE >= 500) &&
               ! (_POSIX_C_SOURCE >= 200112L)
       From glibc 2.12 to 2.19:
           _BSD_SOURCE || _SVID_SOURCE ||
               (_XOPEN_SOURCE >= 500) &&
               ! (_POSIX_C_SOURCE >= 200112L)
       Before glibc 2.12:
           _BSD_SOURCE || _SVID_SOURCE || _XOPEN_SOURCE >= 500

DESCRIPTION      
       brk() and sbrk() change the location of the program break, which
       defines the end of the process's data segment (i.e., the program
       break is the first location after the end of the uninitialized data
       segment).  Increasing the program break has the effect of allocating
       memory to the process; decreasing the break deallocates memory.

   brk() sets the end of the data segment to the value specified by
   addr, when that value is reasonable, the system has enough memory,
   and the process does not exceed its maximum data size (see
   setrlimit(2)).

RETURN VALUE    
       On success, brk() returns zero.  On error, -1 is returned, and errno
       is set to ENOMEM.

还有一点我不明白：如果brk()只是将数据段的末尾设置为addr指定的值，那么为什么它的参数是void*而不是int类型？

感谢您的帮助！

Answer 1

我想，自 "break is the address of the first location beyond the current end of the data region" (Wikipedia) 以来，您不能也不应该知道那里有什么数据，因此指针不能是除 void 之外的任何其他类型。

为什么函数 brk() 的参数是 void* 而不是 int 类型？

why do the argument of function brk() is void* and not int type?

linux

system-calls

brk