How is memory allocated on heap without a system call?
I was wondering that if the space required on heap is not large enough such that there is no need for a brk/sbrk system all (to shift the break pointer (brk) of data segment), how does a library function (such as malloc) allocates space on heap. I am not asking about the data-structures and algorithms for heap management. I am just asking how does malloc get the address of the first location of the heap if it doesn't invoke a system call. I am asking this because I have heard that it is not always necessary to invoke a system call (brk/sbrk) as these are only required to expand the space.Please correct me if I am wrong.
The basic idea is that when your program starts, the heap is very small, but not necessarily zero. If you only allocate (malloc) a small amount of memory, the library is able to handle it within the small amount of space it has when it is loaded. However, when malloc runs out of that space, it needs to make a system call to get more memory.
That system call is often sbrk(), which moves the top of the heap's memory region up by a certain amount. Usually, the malloc library routine increases the heap by larger than what is needed for the current allocation, with the hope that future allocations can be performed w/o making a system call.
Other implementations of malloc use mmap() instead -- this allows the program to create a sparse virtual memory mapping. However, mmap() based malloc implementations do the same thing as the sbrk()-based ones: each system call reserves more memory than what is necessarily needed for the current call.
One way to look at this is to trace a program that uses malloc: you'll see that for N calls to malloc, you will see M system calls (where M is much smaller than N).
The short answer is that it uses sbrk() to allocate a big hunk, which at that point belongs to your app process. It can then further parcel out subsections of that as individual malloc calls without needing to ask the system for anything, until it exhausts that space and needs to resort sbrk() again.
You said you didn't want the details on the data structures, but suffice it to say that the implementation of malloc (ie your own process, not the OS kernel) is keeping track of which space in the region it got from the system is spoken for and which is still available to dole out as individual mallocs. It's like buying a big tract of land, then subdividing it into lots for individual houses.
使用sbrk（）或mmap（） - http://linux.die.net/man/2/sbrk，http://linux.die.net/man/2/mmap
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