Lecture 06: Memory Allocation and Program Layout
Table of Contents
1 Memory Model
So far in programming C, we haven't given a lot of thought to the variables we declare and what it actually means to declare a variable of a given type. Recall that in C the notion of a type and the amount of memory to store that type are strongly linked. For the basic types we've looked at so far, here are their memory requirements:
int
: integer number : 4-bytesshort
: integer number : 2-byteslong
: integer number : 8-byteschar
: character : 1-bytefloat
: floating point number : 4-bytesdouble
: floating point number : 8-bytesvoid *
: pointers : 8-bytes on (64 bit machines)
But, what does it mean for a type to require memory, and where does that memory come from and how is it managed? Understanding the memory model in C is vital to becoming a good programmer because there are situations where you have to use complex memory management to write effective programs. Simple mistakes can lead to programs with mysterious bugs that fail in inexplicable ways.
2 Local Memory Allocation on the Stack
When you declare a variable, you are actually stating to C that you need to create space for the data of that variable to exist. Consider a very simple example.
int a = 10;
The declaration of the integer a
will allocate memory for the
storage for an integer (4-bytes). We refer to the data stored in
memory via the variable a
.
Memory allocation refers to the process by which the program makes "space" for the storage of data. When you declare a variable of a type, enough memory is allocation locally to store data of that type. The allocation is local, occurring within the scope of the function, and when that function returns the memory is deallocated. This should be intuitive based on your experience with programming so far, you can't referenced a variable outside the scope of your function.
However with pointers in C, it's easy to make mistakes where your
pointer references a memory address out of scope of the current
function or even completely unallocated memory. As an example of a
common mistake, consider the simple program below which has a
function plus()
which adds two numbers and returns a memory
address for the result value.
int * plus(int a, int b){ int c = a + b; return &c; //return a reference to //locally declared c } int main(int argc, char * argv[]){ int * p = plus(1,2); printf("%d\n",*p); //dereference return value //to print an integer }
The function plus is declared to take two integers as arguments and
return a pointer to an integer. Within the body of the function, the
two integers are summed together, and the result in stored in c
, a
variable declared locally within the context of the function. The
function then returns the memory address of c
, which is assigned
to the pointer value p
in main.
What's the problem? The memory of c
is deallocated once the
function returns, and now p
is referencing a memory address which
is unallocated. The print statement, which deferences p
,
following the pointer to the memory address, may fail. The above
code is bad, and you can also follow the reasoning with a stack
diagram.
plus(1,2) return &c printf("%d\n",*p) (main) | (main) | (main) | (main) .---.----. | .---.----. | .---.----. | .---.----. | p | .-+-X | | p | .-+-X | | p | .-+---. | | p | .-+---. '---'----' | '---'----' | '---'----' | | '---'----' | | ------------- | ------------ | | | | (plus) | (plus) | | | | .---.---. | .---.---. | | | | | a | 1 | | | a | 1 | | | | | |---|---| | |---|---| | | | | | b | 2 | | | b | 2 | | | | | |---|---| | |---|---| | | | | | c | 3 | | | c | 3 | <--' | X--' | '---'---' | '-------' c exists locally returning a reference When p is dereferenced in plus() to c assined to p it points to unallocated memory
First, in main()
, p waits for the result of the call to plus()
,
which set's c
. Once plus()
returns, the value of p
references
a variable declared in plus()
, but all locally declared variables
in plus()
are deallocated once plus()
returns. That means, by
the time the printf()
is called and p
is dereferenced, the
memory address references unallocated memory, and we cannot
guarantee that the data at that memory address will be what we
expect.
2.1 The Stack
Another term for local memory allocation is stack allocation because the way programs track execution across functions is based on a stack. A stack is a standard ordred data structure, like a list, that has the property that the last item inserted on the stack is the first item that is removed. This is often referred to as LIFO data structure, last-in-first-out. A stack has two primary functions:
- push : push an item on to the top of the stack
- pop : pop an item off the top of the stack
The stack's top always references the last item pushed onto the stack, and the items below the top are ordered base on when they were pushed on. The most recently pushed items come first. This means when you pop items off the top stack, the next item becomes the next top, which maintains the LIFO principle.
The stack model (last-in-first-out) matches closely the model of function calls and returns during program execution. The act of calling a function is the same as pushing the called function execution onto the top of the stack, and, once that function completes, the results are returned popping the function off the stack.
Each function is contained within a structure on the stack called a stack frame. A stack frame contains all the allocated memory from variable deliberations as well as a pointer to the execution point of the calling function, the so called return pointer. A very common exploit in computer security is a buffer overflow attack where an attacker overwrite the return pointer such that when the function returns, code chosen by the attacker is executed.
To understand how function calls are modeled in the stack, we have
nested function calls under addonetwo()
, and which ever function
is currently executing has the stack frame at the top of the stack
and the calling function, where to the current function returns, is
the stack frame next from top. When the current function returns,
its stack frame is popped off the stack, and the calling function,
now the top of the stack, continues executing from the point of
function call.
int gettwo(){ return 2; } int getone(){ return 1; } int addonetwo(){ int one = getone(); int two = gettwo(); return one+two; } int main(){ int a = func2(); }
program executed .------ top of stack v main() push main | | main() | | '----------------' addonetwo() | push addonetwo | | addonetwo() | | | main() | | '----------------' getone() | | | getone() | push getone | | addonetwo() | | | main() | | '----------------' return 1 pop | | addonetwo() | | | main() | | '----------------' gettwo() | | | gettwo() | push gettwo | | addonetwo() | | | main() | | '----------------' return 2 pop | | addonetwo() | | | main() | | '----------------' return 1 + 2 pop | | main() | | '----------------' program exits
The act of pushing and popping functions onto the stack also affects the memory allocation. By pushing a function onto the stack, the computer is actually allocating memory for the function's local variables, and once that function returns, the function and its allocated memory is popped off the stack, deallocating it. This is why local declared variables are also called stacked variables.
Following the example from before we can now better understand why it fails by adding in the pushes and pops of the stack.
*PUSH* *POP* plus(1,2) return &c printf("%d\n",*p) (main) | (main) | (main) | (main) .---.----. | .---.----. | .---.----. | .---.----. | p | .-+-X | | p | .-+-X | | p | .-+---. | | p | .-+---. '---'----' | '---'----' | '---'----' | | '---'----' | | ------------- | ------------ | | | | (plus) | (plus) | | | | .---.---. | .---.---. | | | | | a | 1 | | | a | 1 | | | | | |---|---| | |---|---| | | | | | b | 2 | | | b | 2 | | | | | |---|---| | |---|---| | | | | | c | 3 | | | c | 3 | <--' | X--' | '---'---' | '-------' Pusing plus() The return of plus() When p is dereferenced onto the stack pops it off the stack in the print, p now allocates memory unallocated all stack references unallocated for c variables including c memory
3 Global Memory Allocation on the Heap
Just because the sample program with plus()
from the previous
section doesn't work properly when returning a memory reference, it
does not mean you cannot write functions that return a memory
reference. What is needed is a different allocation procedure for
global memory which is not deallocated automatically when functions
return and thus remains in scope for the entirety of the program
execution.
In fact, you have already seen how to do this in C++ when you using
the new
construct. When you call new for some structure, what
is actually going on? Consider the small code snippet below:
Node * node = new Node();
The local variable declaration is for the variable node
, but
that's just a pointer to some memory. The variable node
, itself,
is declared on the stack and has enough memory to store a memory
address. The value of that memory address is set by the return of
the call new Node()
. The new
function will automatically
allocate enough memory to store a Node
structure and the node
variable now references that memory. But, where does this new
memory come from?
If you think about it, the memory cannot have been allocated on the
stack because it is memory being returned from a function, the
new
function. As we've seen previously, if a function returns a
local reference of a variable declared on the stack, that memory is
automatically deallocated when the function returns. Instead, this
memory must have been allocated somewhere else, and it is. The
new
function performs a dynamic memory allocation in global
memory that is not associated with scope of functions or the
stack. It is instead allocating on the heap.
3.1 The heap, malloc()
and free()
The global memory region for a program is called the heap, which is a fragmented data structure where new allocation try and fit within unallocated regions. Whenever a program needs to allocate memory globally or in a dynamic way, that memory is allocated on the heap, which is shared across the entire program irrespective of function calls.
In C the new
function is called malloc()
, or memory
allocator. The malloc()
function takes the number of bytes to be
allocated as its argument, and it returns a pointer to a memory
region on the heap of the requested byte-size. Here is a code
snippet to allocate memory to store an integer on the heap:
// .--- Allocate sizeof(int) number of bytes // v int * p = (int *) malloc(sizeof(int)); // ^ // '-- Cast to a integer pointer
First, to allocate an integer on the heap, we have to know how
big an integer is, that is, what size is it, which we learn via
sizeof()
macro. Since an int
is 4 byes in size, malloc()
will
allocate 4 bytes of memory on the heap in which an integer can be
stored. malloc()
then returns the memory address of the newly
allocated memory, which is assigned to p
. Since malloc()
is a
general purpose allocation tool, just allocating bytes which can be
used to store data generally, we have to cast the resulting pointer
value to the right type, int *
. If we don't, the program won't
fail, but you will get a compiler warning.
We can now use p
like a standard pointer as before; however, once
we're done with p
we have to explicitly deallocate it. Unlike
stack based memory allocations which are implicitly deallocated
when functions return, there is no way for C to know when you are
done using memory allocated on the heap. C does not track
references, like Java, so it can't perform garbage collection;
instead, you, the programmer, must indicate when you're done with
the memory by freeing. The deallocation function is free()
(equivalent to delete
in C++), which takes a pointer value as
input and "frees" the referenced memory on the heap.
int * p = (int *) malloc(sizeof(int)); //do something with p free(p); //<-- deallocate p
With all of that, we can now complete the plus()
program to
properly return a memory reference to the result.
int * plus(int a, int b){ int *p = (int *) malloc(sizeof(int)); //allocate enough space for *p = a + b; //for an integer return p; //return pointer } int main(int argc, char * argv[]){ int * p = plus(1,2); //p now references memory on the heap printf("%d\n",*p); free(p); //free allocated memory }
3.2 Memory Leaks and other Memory Violations
In C (and C++), the programer is responsible for memory management, which includes both the allocation and deallocation of memory. As a result, there are many mistakes that can be made, which is natural considering that all programers make mistakes. Perhaps the most common mistake is a memory leak, where heap allocated memory is not freed. Consider the following program.
int main(int argc, char * argv[]){ int i, * p; for(i=0;i>100;i++){ p = (int *) malloc(sizeof(int)); *p = i; } }
At the malloc, on Line 5, the returned pointer to newly
allocated memory is overwriting the previous value of p
. There is
no free()
occuring, and once the previous pointer value is
overwritten, there is no way to free that memory. It is considered
lost, and the above program has a memory leak. Memory leaks are
very bad, and over time, can cause your program to fail.
Another common mistake is dereferencing a dangling pointer. A
dangling poiner is when a pointer value once referenced allocated
memory, but that memory has seen been dealocated. We'e seen an
example of this already in the plus()
program, but it can also
occur for heap allocations.
int main(int argc, char * argv[]){ int *p = (int *) malloc(sizeof(int)); //... code free(p); //... code *p = 10; }
Once p
has been freed, the memory referenced by the p
's value
can be reclaimed by other allocations. At the point where p
is
dereferenced for the assignment, it might be the case that you are
actually overwriting memory for some other value, and corrupting
your program. Once memory is freed, it should never be
dereferenced. These kinds of memory violations can lead to the
dreaded SEGFAULT
.
Another, common mistake with memory allocation is a double
free. The heap allocation functions maintain special data
structures so that it is easy to find unallocated memory and
reallocate for future malloc()
calls. If you call free()
twice
on a pointer, you will corrupt that process, result in a core
dump
or some other very scary error.
4 Program Layout: Stack vs. Heap
Now that you understand the two different memory allocation procedures, let's zoom out and take a larger look at how memory in programs is managed more generally. Where is the stack? Where is the heap? How do they grow or shrink?
To answer these questions, you first need to think of a program as a memory profile. All information about a program, including the actual binary code and variables all are within the memory layout of a program. When executing, the Operating System will manage that memory layout, and a snapshot of that memory and the current execution point basically defines a program. This allows the operating system to swap in and out programs as needed.
On 64-bit machines, the total available memory addresses are from 0 to 264-1. For a program, the top and bottom of the address space are what is important. We can look at the program's memory layout in diagram form:
2^64-1---> .----------------------. High Addresses | Enviroment | |----------------------| | | Functions and variable are declared | STACK | on the stack. base pointer -> | - - - - - - - - - - -| | | | | v | : : . . The stack grows down into unused space . Empty . while the heap grows up. . . . . : : | ^ | | | | break point -> | - - - - - - - - - - -| Dynamic memory is declared on the heap | HEAP | | | |----------------------| | BSS | The compiled binary code is down here as |----------------------| well as static and initialzed data | Data | |----------------------| Low Addresses | Text | 0 -----> '----------------------'
At the higher addresses is the stack and at the lower address is the heap. The two memory allocation regions grow into the middle of the address space, which is unused and unallocated. In this way, the two allocations will not interfere with each other. The stack base pointer is the current top of the stack, and as functions are called and returned, it will shift appropriately. The break point refers to the top of the programs data segment, which contains the heap. As the heap fills up, requirement more space, the break is set to higher addresses.
You should note that this memory layout is virtual. From the program's perspective it has access to the entire address range, but in reality, this might not be the case because the program is sharing physical memory with other programs, including the operating system. How that process works is a discussion for another class.
4.1 Memory Mapping and Dynamic Libraries
Between the break point and the base pointer is unallocated memory, but it may not be unused. This region can be memory mapped, which is the process of loading data directly from files into memory. As the programmer, you can directly memory map files, but often this is done automatically for you when you read and write files.
Another common use for the middle addresses is the loading of
dynamic shared libraries. When you make a call to a function like
printf()
or malloc()
, the code for those functions exist in
shared libraries, the standard C library, to be precise. Your
program must run that code, but the operating system doesn't want to
have load more code into memory than needed. Instead, the O.S. loads
shared libraries dynamically, and when it does so, it maps the
necessary code into the middle address spaces.
4.2 Address Space Randomization
A final note about program memory address is that it is not consistent across runs. Take a look at the very simple program:
#+NAME random.c
#include <stdio.h> #include <stdlib.h> int main(int argc, char * argv[]){ int a; printf("0x%p\n",&a); //print the address of a }
All this program does is print the address of the statically
declared variable a
. With your knowledge of the memory address
layout, you could probably say with confidence that the address
value should be a relatively large number, but you might also
expect that it would be consistent across runs of the program.
#> for i in `seq 1 1 10`; do ./random; done 0x0x7fff58e0cafc 0x0x7fff50fffafc 0x0x7fff59a6dafc 0x0x7fff508c3afc 0x0x7fff533c8afc 0x0x7fff539a9afc 0x0x7fff528c6afc 0x0x7fff50984afc 0x0x7fff52f13afc 0x0x7fff5c97aafc
In fact, it is not consistent. While the addresses are relatively high in the address space, there least significant bits are quite random. This is on purpose, and for security purposes. The stack values also contain important pointers to next bit of code that should be executed, this is how program knows how to return to the previous execution prior to the function call. In a buffer overflow attack, the attacker overwrites the return pointer allowing code to be executed of the attacker's choosing, which can be exploit code. In order for the attack to be successful, the attacker must know exactly where in the program to set the return pointer, but if the addresses are randomized, this because a much more difficult tasks. Which is why the address space is randomized.