CODEWITHSUNDEEP
cpointers
Hanan
Hanan

Reputation: 1179

how to assign pointer value of another pointer

void funcF(char *outBuffer)
{
        char * inBuffer;
        inBuffer = (char *) malloc(500);
        // add stuff to inBuffer
        strcpy(inBuffer, "blabla");

        outBuffer = inBuffer; //probably this is wrong 
}

int main()
{
    char * outBuffer;

    funcF(outBuffer);

    printf("%s", outBuffer); // i want to get "blabla" as output
    free(outBuffer);
}

My question how can i make outBuffer point to the same address as inBuffer so that i can access the data in inBuffer ?

Upvotes: 0

Views: 294

Answers (3)

Nick
Nick

Reputation: 10539

Not sure if this is what you want, but:

int main() {

const unsigned int MAX_BUFF = 1024

char outBuffer[MAX_BUFF];

funcF(outBuffer);

printf("%s", outBuffer); // i want to get "blabla" as output

/* free(outBuffer); */

}

will have problem with free(), but you get the idea.

Upvotes: -2

simonc
simonc

Reputation: 42195

Your current code passes a pointer by value. This means that funcF operates on a copy of the caller's pointer. If you want to modify the caller's pointer, you need to either pass the address of that pointer (i.e. a pointer to a pointer):

void funcF(char **outBuffer)
{
    char * inBuffer = malloc(500);
    strcpy(inBuffer, "blabla");
    *outBuffer = inBuffer;
}

int main()
{
    char * outBuffer;
    funcF(&outBuffer);
    //    ^

or change funcF to return a pointer:

char* funcF()
{
    char* inBuffer = malloc(500);
    strcpy(inBuffer, "blabla");
    return inBuffer;
}

int main()
{
    char * outBuffer = funcF();

Upvotes: 4

Shafik Yaghmour
Shafik Yaghmour

Reputation: 158599

You need to pass a char **:

void funcF(char **outBuffer)

then assign like this:

*outBuffer = inBuffer;

and pass it in like so:

funcF(&outBuffer);

you can alternatively, have it return a char *.

Upvotes: 2

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