Allocate memory inside function ??

fx9200

the code is too long but : #define STRUCT struct mystruct void function (void *input){ .... pointer = (STRUCT*) malloc (sizeof(STRUCT)); //filling in input attributes... input = pointer; } main(){ void *ptr; function (ptr); printf("%...", ptr->..........); } this work only if i allocate the memory for ptr in the main program

toxcct

fx9200 wrote:

pointer = (STRUCT*) malloc (sizeof(STRUCT));

why don't you directly write into input ?

input = (STRUCT*) malloc (sizeof(STRUCT));

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fx9200

i tried it i had the same result in all cases

prasad_som

Change this to,

void function (void **input){
....
*pointer = (STRUCT*) malloc (sizeof(STRUCT));

//filling in input attributes...

input = pointer;
}

main(){
void *ptr;
function (&ptr);
printf("%...", ptr->..........);
}

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fx9200

how to pass reference(pointer) by reference ? please give example

toxcct

he seems to doing C, not C++... so no reference available (in the sense of C++ understands it). but pointers remain good

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fx9200

:rolleyes: yes so ?

toxcct

fx9200 wrote:

yes so ?

so, prasad_som[^] gave you an answer... hve you read it ?

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fx9200

I tried this it works , but i cannot fix the allocated emory size by a no-static way void fill(void *input); struct mystruct { char *ch; int n; }; #define ARG_SIZE sizeof(struct mystruct) int main(){ void *ptr; ptr = (void*) malloc (ARG_SIZE); fill(ptr); printf ("%i",((struct mystruct*)ptr)->n); printf ("--%s",((struct mystruct*)ptr) ->ch); return 0; } void fill(void *input){ struct mystruct* st; st = (struct mystruct*) input; st ->n = 3;//operation is ok st->ch = (char *) malloc (3*sizeof(char)); strcpy(st->ch,"OK"); input = (void*) st; }

prasad_som

Though, its not applicable in C context. You need to understand theory behind Cedric's suggesstion. In you original code, you was passing pointer argument. Though its a pointer, it will be passed by value, and inside that function, memory allocated will be not at address you expected. In that case you need to pass pointer to pointer, as shown in my previous reply.

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fx9200

i'm trying thanks

fx9200

but how the code below works; i allocated memory for chr* inside function

prasad_som

Which code you are talking about ?

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fx9200

void fill(void *input); struct mystruct { char *ch; int n; }; #define ARG_SIZE sizeof(struct mystruct) int main(){ void *ptr; ptr = (void*) malloc (ARG_SIZE); fill(ptr); printf ("%i",((struct mystruct*)ptr)->n); printf ("--%s",((struct mystruct*)ptr) ->ch); return 0; } void fill(void *input){ struct mystruct* st; st = (struct mystruct*) input; st ->n = 3;//operation is ok st->ch = (char *) malloc (3*sizeof(char)); strcpy(st->ch,"OK"); input = (void*) st;

fx9200

void fill(void *input); struct mystruct { char *ch; int n; }; #define ARG_SIZE sizeof(struct mystruct) int main(){ void *ptr; ptr = (void*) malloc (ARG_SIZE); fill(ptr); printf ("%i",((struct mystruct*)ptr)->n); printf ("--%s",((struct mystruct*)ptr) ->ch); return 0; } void fill(void *input){ struct mystruct* st; st = (struct mystruct*) input; st ->n = 3;//operation is ok st->ch = (char *) malloc (3*sizeof(char)); strcpy(st->ch,"OK"); input = (void*) st; }

prasad_som

Here, though you are allocating memory for ch member of mystruct inside the function, you have passed mystruct* as parameter and , for that memory is already allocated, before calling this function. And bitwise(shallow) copy happens in this case. p.s. Dont's create multiple threads for same code.

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fx9200

ok thanks

Roger Stoltz

What I think you need to understand after having read your previous posts, is that when you call a function the arguments passed are copies on the stack. Have a look here[^] if you're not sure what the stack is and how it is used. Let's examine this with some code examples. A function called AddOne is used for increasing a variable by the value of 1.

void AddOne( int nTheVariable )
{
    nTheVariable = nTheVariable + 1;
}

void main()
{
    int nMyValue = 0;
    printf( "The value is %d.\n", nMyValue ); // Prints "The value is 0."
    AddOne( nMyValue );
    printf( "The value is %d.\n", nMyValue ); // Also prints "The value is 0."
}

When AddOne() is declared as above, the value of nMyValue is copied to the stack and read by AddOne(). When the function returns is has indeed increased its local copy (nTheVariable) by one, but nMyValue still remains the same since it was a copy of it that was passed to the function. If you want to alter the value of nMyValue, you have to pass its location as argument to AddOne(). "Location" in this aspect means "the address of". In code it would look like this:

void AddOne( int* pnTheVariable )
{
    // Here we've got a copy of the address of the variable so
    // we can modify it at its original location 
    *pnTheVariable = *pnTheVariable + 1;
}

void main()
{
    int nMyValue = 0;
    printf( "The value is %d.\n", nMyValue ); // Prints "The value is 0."
    AddOne( &nMyValue ); // Pass a copy of the address of nMyValue
    printf( "The value is %d.\n", nMyValue ); // Prints "The value is 1."
}

In your case you want to allocate memory and you always assign it to a pointer. To be able to alter the address assigned to a pointer, you have to pass a copy of the address of the pointer to the function. This is still the same as above, but now you want to alter the value of a pointer. In code it would look something like this if you want to allocate memory for three ints:

void Allocate( int** ppnTheMemory )
{
    *ppTheMemory = (int*)malloc( 3 * sizeof( int ) );
}

void main()
{
    int* pnMyValues = NULL;
    Allocate( &pnMyValue ); // Pass the address of the pointer
    if( pnMyValues )
    {
        free( pnMyValues );

David Crow

fx9200 wrote:

how to pass reference(pointer) by reference ? please give example

See here.

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