About the memory allocated for a string variable in C

Question

I'm writing a function that receives a string and extracts the tokens from it, storing them in a stack. There is a variable called currentToken, which starts with memory for only 1 character:

char *currentToken = ( char * )malloc( sizeof( char ) );

As the token is being formed, currentToken is extended through realloc to accomodate the new character. Everytime currentToken is complete, it is added to the stack by reference. Then, I try to "reset" it (as if I was setting it to an empty string) freeing its memory and allocating it again. Does it destroy the data previously included in the stack? If so, how do I fix this problem? Thanks in advance.

The stack is implemented as a structure, and it is initialized right from the beginning:

typedef struct stackOfStrings {
    char **array;
    int numberOfElements;
} StackOfStrings;

/* Initializes the stack of strings: */
void initializeStackOfStrings( StackOfStrings *sPtr )
{
    sPtr->numberOfElements = 0;
    sPtr->array = ( char ** )malloc( 1 * sizeof( char * ) );
}

/* Inserts str at the top the stack of strings, returning 1 if it succeeds, or 0
otherwise: */
int pushOnStackOfStrings( StackOfStrings *sPtr, char *string )
{
    int length = string_length( string );

    ++( sPtr->numberOfElements );

    sPtr->array = realloc( sPtr->array, ( sPtr->numberOfElements ) * sizeof( char * ) );

    if ( sPtr->array == NULL )
    {
        return 0;
    }

    *( sPtr->array + ( sPtr->numberOfElements - 1 ) ) = ( char * )malloc( length * sizeof( char ) );

*( sPtr->array + ( sPtr->numberOfElements - 1 ) ) = string;

return 1;

}

You are losing memory. Last 2 lines, you are allocating some memory, then lose the pointer to that memory by assigning `string` to it. — Xaqq, Feb 08 '14 at 21:40
But then how would I copy the string to that part? Should I use memcpy? — Fernando Karpinski, Feb 08 '14 at 21:41
If it's a proper C-string (i.e. null-terminated) [`strcpy`](http://en.cppreference.com/w/c/string/byte/strcpy) is easier, otherwise use `memcpy`. — Kninnug, Feb 08 '14 at 21:50
@glglgl, if you don't cast the `malloc()` you will get a warning. — Stephen Rasku, Feb 08 '14 at 22:00
You could use `strdup()` as well. It will do the allocation and copy as one operation. You will have to free the memory as it is dynamically allocating memory. — Stephen Rasku, Feb 08 '14 at 22:10
@StephenRasku Why would that be a warning? Casting return value of malloc is not the done thing. — David Heffernan, Feb 08 '14 at 22:14
If you get a warning when you fail to cast the return from `malloc()`, then you're either using the wrong language, or using the language wrong. — Crowman, Feb 08 '14 at 23:52
@StephenRasku A warning is normally not supposed to be suppressed with a cast, but dealt with appropriately, e. g. by `#include `. See [here](http://stackoverflow.com/q/605845/296974). — glglgl, Feb 09 '14 at 07:43
@glglgl, malloc returns a `void *` not a `char *`. `#include ` won't help. — Stephen Rasku, Feb 11 '14 at 02:02
@StephenRasku I don't know what you mean. A `void *` is compatible with all data pointers. (We are still talking about C, aren't we?) — glglgl, Feb 11 '14 at 07:50

score 0 · Answer 1 · answered Feb 09 '14 at 01:40

try this

void initializeStackOfStrings( StackOfStrings *sPtr )
{
    sPtr->numberOfElements = 0;
    sPtr->array = NULL;
}
...
sPtr->array[sPtr->numberOfElements - 1] = ( char * )malloc( (length+1) * sizeof( char ) );
strcpy(sPtr->array[sPtr->numberOfElements - 1], string);

About the memory allocated for a string variable in C

1 Answers1