Where does malloc() allocate memory? Is it the data section or the heap section of the virtual address space of the process?

Question

Ever since I was introduced to C, I was told that in C dynamic memory allocation is done using the functions in the malloc family. I also learned that memory dynamically allocated using malloc is allocated on the heap section of the process.

Various OS textbooks say that malloc involves system call (though not always but at times) to allocate structures on heap to the process. Now supposing that malloc returns pointer to chunk of bytes allocated on the heap, why should it need a system call. The activation records of a function are placed in the stack section of the process and since the "stack section" is already a part of the virtual address space of the process, pushing and popping of activation records, manipulation of stack pointers, just start from the highest possible address of the virtual address space. It does not even require a system call.

Now on the same grounds since the "heap section" is also a part of the virtual address space of the process, why should a system call be necessary for allocating a chunk of bytes in this section. The routine like malloc could self handle the "free" list and "allocated" list on its own. All it needs to know is the end of the "data section". Certain texts say that system calls are necessary to "attach memory to the process for dynamic memory allocation", but if malloc allocates memory on "heap section" why is it at all required to attach memory to the process during malloc? Could be simply taken from portion already part of the process.

While going through the text "The C Programming Language" [2e] by Kernighan and Ritchie, I came across their implementation of the malloc function [section 8.7 pages 185-189]. The authors say :

malloc calls upon the operating system to obtain more memory as necessary.

Which is what the OS texts say, but counter intuitive to my thought above (if malloc allocates space on heap).

Since asking the system for memory is a comparatively expensive operation, the authors do not do that on every call to malloc, so they create a function morecore which requests at least NALLOC units; this larger block is chopped up as needed. And the basic free list management is done by free.

But the thing is that the authors use sbrk() to ask the operating system for memory in morecore. Now Wikipedia says:

brk and sbrk are basic memory management system calls used in Unix and Unix-like operating systems to control the amount of memory allocated to the data segment of the process.

Where

a data segment (often denoted .data) is a portion of an object file or the corresponding address space of a program that contains initialized static variables, that is, global variables and static local variables.

Which I guess is not the "heap section". [Data section is the second section from bottom in the picture above, while heap is the third section from bottom.]

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I am totally confused. I want to know what really happens and how both the concepts are correct? Please help me understand the concept by joining the scattered pieces together...

Wacker

The heap is part of the data section. – Staysail

malloc() isn't a system call. – Staysail

@Barmar: No, but malloc makes an OS call to obtain the memory being requested. – Fill

Why would you care? malloc() returns an address, and that is exactly what you need. BTW: you can ignore physical memory; in practice user processes only have to deal with (virtual) addresses. – Butterball

Abhishek Ghosh, your malloc() understanding is only one of many ways allocation is handled. A clear understanding of one model may hinder (or help) understanding of the many possibiitlies. – Ethelinda

Answer 1

In your diagram, the section labeled "data" is more precisely called "static data"; the compiler pre-allocates this memory for all the global variables when the process starts.

The heap that malloc() uses is the rest of the process's data segment. This initially has very little memory assigned to it in the process. If malloc() needs more memory, it can use sbrk() to extend the size of the data segment, or it can use mmap() to create additional memory segments elsewhere in the address space.

Why does malloc() need to do this? Why not simply make the entire address space available for it to use? There are historical and practical reasons for this.

The historical reason is that early computers didn't have virtual memory. All the memory assigned to a process was swapped in bulk to disk when switching between processes. So it was important to only assign memory pages that were actually needed.

The practical reason is that this is useful for detecting various kinds of errors. If you've ever gotten a segmentation violation error because you dereferenced an uninitialized pointer, you've benefited from this. Much of the process's virtual address space is not allocated to the process, which makes it likely that unitialized pointers point to unavailable memory, and you get an error trying to use it.

There's also an unallocated gap between the heap (growing upwards) and the stack (growing downward). This is used to detect stack overflow -- when the stack tries to use memory in that gap, it gets a fault that's translated to the stack overflow signal.

Staysail

Answer 2

This is the Standard C Library specification for malloc(), in its entirety:

7.22.3.4 The malloc function

Synopsis

#include <stdlib.h>
void *malloc(size_t size);

Description

The malloc function allocates space for an object whose size is specified by size and whose value is indeterminate. Note that this need not be the same as the representation of floating-point zero or a null pointer constant.

Returns

The malloc function returns either a null pointer or a pointer to the allocated space.

That's it. There's no mention of the Heap, the Stack or any other memory location, which means that the underlying mechanisms for obtaining the requested memory are implementation details.

In other words, you don't care where the memory comes from, from a C perspective. A conforming implementation is free to implement malloc() in any way it sees fit, so long as it conforms to the above specification.

Fill

But isn't the OP allowed to be curious about how it typically works inside? – Debbradebby

The OP already knows how it works inside. – Fill

Answer 3

I was told that in C dynamic memory allocation is done using the functions in the malloc family. I also learned that memory dynamically allocated using malloc is allocated on the heap section of the process.

Correct on both points.

Now supposing that malloc returns pointer to chunk of bytes allocated on the heap, why should it need a system call.

It needs to request an adjustment to the size of the heap, to make it bigger.

...the "stack section" is already a part of the virtual address space of the process, pushing and popping of activation records, manipulation of stack pointers, [...] does not even require a system call.

The stack segment is grown implicitly, yes, but that's a special feature of the stack segment. There's typically no such implicit growing of the data segment. (Note, too, that the implicit growing of the stack segment isn't perfect, as witness the number of people who post questions to SO asking why their programs crash when they allocate huge arrays as local variables.)

Now on the same grounds since the "heap section" is also a part of the virtual address space of the process, why should a system call be necessary for allocating a chunk of bytes in this section.

Answer 1: because it's always been that way.
Answer 2: because you want accidental stray pointer references to crash, not to implicitly allocate memory.

malloc calls upon the operating system to obtain more memory as necessary.

Which is what the OS texts say, but counter intuitive to my thought above (if malloc allocates space on heap).

Again, malloc does request space on the heap, but it must use an explicit system call to do so.

But the thing is that the authors use sbrk() to ask the operating system for memory in morecore. Now Wikipedia says:

brk and sbrk are basic memory management system calls used in Unix and Unix-like operating systems to control the amount of memory allocated to the data segment of the process.

Different people use different nomenclatures for the different segments. There's not much of a distinction between the "data" and "heap" segments. You can think of the heap as a separate segment, or you can think of those system calls -- the ones that "allocate space on the heap" -- as simply making the data segment bigger. That's the nomenclature the Wikipedia article is using.

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Some updates:

I said that "There's not much of a distinction between the 'data' and 'heap' segments." I suggested that you could think of them as subparts of a single, more generic data segment. And actually there are three subparts: initialized data, uninitialized data or "bss", and the heap. Initialized data has initial values that are explicitly copied out of the program file. Uninitialized data starts out as all bits zero, and so does not need to be stored in the program file; all the program file says is how many bytes of uninitialized data it needs. And then there's the heap, which can be thought of as a dynamic extension of the data segment, which starts out with a size of 0 but may be dynamically adjusted at runtime via calls to brk and sbrk.

I said, "you want accidental stray pointer references to crash, not to implicitly allocate memory", and you asked about this. This was in response to your supposition that explicit calls to brk or sbrk ought not to be required to adjust the size of the heap, and your suggestion that the heap could grow automatically, implicitly, just like the stack does. But how would that work, really?

The way automatic stack allocation works is that as the stack pointer grows (typically "downward"), it eventually reaches a point that it points to unallocated memory -- that blue section in the middle of the picture you posted. At that point, your program literally gets the equivalent of a "segmentation violation". But the operating system notices that the violation involves an address just below the existing stack, so instead of killing your program on an actual segmentation violation, it quick-quick makes the stack segment a little bigger, and lets your program proceed as if nothing had happened.

So I think your question was, why not have the upward-growing heap segment work the same way? And I suppose an operating system could be written that worked that way, but most people would say it was a bad idea.

I said that in the stack-growing case, the operating system notices that the violation involves an address "just below" the existing stack, and decides to grow the stack at that point. There's a definition of "just below", and I'm not sure what it is, but these days I think it's typically a few tens or hundreds of kilobytes. You can find out by writing a program that allocates a local variable

char big_stack_array[100000];

and seeing if your program crashes.

Now, sometimes a stray pointer reference -- that would otherwise cause a segmentation violation style crash -- is just the result of the stack normally growing. But sometimes it's a result of a program doing something stupid, like the common error of writing

char *retbuf;
printf("type something:\n");
fgets(retbuf, 100, stdin);

And the conventional wisdom is that you do not want to (that is, the operating system does not want to) coddle a broken program like this by automatically allocating memory for it (at whatever random spot in the address space the uninitialized retbuf pointer seems to point) to make it seem to work.

If the heap were set up to grow automatically, the OS would presumably define an analogous threshold of "close enough" to the existing heap segment. Apparently stray pointer references within that region would cause the heap to automatically grow, while references beyond that (farther into the blue region) would crash as before. That threshold would probably have to be bigger than the threshold governing automatic stack growth. malloc would have to be written to make sure not to try to grow the heap by more than that amount. And true, stray pointer references -- that is, program bugs -- that happened to reference unallocated memory in that zone would not be caught. (Which is, it's true, what can happen for buggy, stray pointer references just off the end of the stack today.)

But, really, it's not hard for malloc to keep track of things, and explicitly call sbrk when it needs to. The cost of requiring explicit allocation is small, and the cost of allowing automatic allocation -- that is, the cost of the stray pointer bugs not caught -- would be larger. This is a different set of tradeoffs than for the stack growth case, where an explicit test to see if the stack needed growing -- a test which would have to occur on every function call -- would be significantly expensive.

Finally, one more complication. The picture of the virtual memory layout that you posted -- with its nice little stack, heap, data, and text segments -- is a simple and perhaps outdated one. These days I believe things can be a lot more complicated. As @chux wrote in a comment, "your malloc() understanding is only one of many ways allocation is handled. A clear understanding of one model may hinder (or help) understanding of the many possibilities." Among those complicating possibilities are:

• A program may have multiple stack segments maintaining multiple stacks, if it supports coroutines or multithreading.
• The mmap and shm_open system calls may cause additional memory segments to be allocated, scattered anywhere within that blue region between the heap and the stack.
• For large allocations, malloc may use mmap rather than sbrk to get memory from the OS, since it turns out this can be advantageous.

See also Why does malloc() call mmap() and brk() interchangeably?

As the bard said, "There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy." :-)

Debbradebby

Thanks a lot for the answer. :) Just one request, could you please explain me the meaning of the statement :because you want accidental stray pointer references to crash, not to implicitly allocate memory. I could not understand it. – Wacker

@AbhishekGhosh barmar addresses this point in his answer, and I have also updated mine. – Debbradebby

Answer 4

• Not all virtual addresses are available at the beginning of a process.
• OS does maintain a virtual-to-physics map, but (at any given time) only some of the virtual addresses are in the map. Reading or Writing to an virtual address that isn't in the map cause a instruction level exception. sbrk puts more addresses in the map.
• Stack just like data section but has a fixed size, and there is no sbrk-like system call to extend it. We can say there is no heap section, but only a fixed-size stack section and a data section which can be grown upward by sbrk.
• The heap section you say is actually a managed (by malloc and free) part of the data section. It's clear that the code relating to heap management is not in OS kernel but in C library executing in CPU user mode.

Apothecium