This was originally a Stackoverflow answer.

Dynamic memory allocation

Is memory allocated at runtime using calloc(), 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;

Technically, extern and static are two separate classes of variables in C.

extern int a; /* Declaration */
int a; /* Definition */

Register Memory

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 :)

References:

Resources