Lines Matching full:sbrk
5831 // We assume that brk()/sbrk() does not initialise new memory. Is this
5834 // 0) sbrk() can *decrease* process address space. No zero fill is done
5838 // new] sbrk(0) should be considered to be uninitialized, but in practice
5844 // operating system. So if sbrk(0) already is page aligned, then
5845 // sbrk(PAGE_SIZE) *does* zero-fill the new memory.
5848 // changed. So if (x = sbrk(0)) is not page aligned, then
5849 // sbrk(PAGE_SIZE) yields ((PAGE_SIZE -1) & -x) bytes which keep their
5851 // ((PAGE_SIZE -1) & x) of them are "covered" by the sbrk(), and the rest
5854 // anything that lives above [the new] sbrk(0) should be considered
5860 // A key property of sbrk/brk is that new whole pages that are supplied
5865 // sbrk and brk are not POSIX. However, any system that is a derivative
5866 // of *nix has sbrk and brk because there are too many softwares (such as
5868 // .data+.bss, stack) and the existence of sbrk/brk.
5876 // [More: I think most of what John said only applies to sbrk(). It seems
5878 // directly with brk(), not with sbrk(), perhaps it would be reasonable to
