This was originally a Stackoverflow answer.
Dynamic memory allocation
Is memory allocated at runtime using
malloc() and friends. It is sometimes also referred to as ‘heap’ memory, although it has nothing to do with the heap data-structure ref.
int * a = malloc(sizeof(int));
Heap memory is persistent until
free() is called. In other words, you control the lifetime of the variable.
Automatic memory allocation
This is what is commonly known as ‘stack’ memory, and is allocated when you enter a new scope (usually when a new function is pushed on the call stack). Once you move out of the scope, the values of automatic memory addresses are undefined, and it is an error to access them.
int a = 43;
Note that scope does not necessarily mean function. Scopes can nest within a function, and the variable will be in-scope only within the block in which it was declared. Note also that where this memory is allocated is not specified. (On a sane system it will be on the stack, or registers for optimisation)
Static memory allocation
Is allocated at compile time, and the lifetime of a variable in static memory is the lifetime of the program.
In C, static memory can be allocated using the
static keyword. The scope is the compilation unit only.
Things get more interesting when the
extern keyword is considered. When an
extern variable is defined the compiler allocates memory for it. When an
extern variable is declared, the compiler requires that the variable be defined elsewhere. Failure to declare/define
extern variables will cause linking problems, while failure to declare/define
static variables will cause compilation problems.
in file scope, the static keyword is optional (outside of a function):
int a = 32;
But not in function scope (inside of a function):
static int a = 32;
static are two separate classes of variables in C.
extern int a; /* Declaration */ int a; /* Definition */
The last memory class are ‘register’ variables. As expected, register variables should be allocated on a CPU’s register, but the decision is actually left to the compiler. You may not turn a register variable into a reference by using address-of.
register int meaning = 42; /* this is wrong and will fail at compile time. */ printf("%p\n",&meaning);
Most modern compilers are smarter than you at picking which variables should be put in registers :)
- The libc manual
- K&R’s The C programming language, Appendix A, Section 4.1, “Storage Class”. (PDF)
- C11 standard, section 5.1.2, 188.8.131.52
- Wikipedia also has good pages on Static Memory allocation, Dynamic Memory Allocation and Automatic memory allocation
- The C Dynamic Memory Allocation page on Wikipedia
- This Memory Management Reference has more details on the underlying implementations for dynamic allocators.