how to assign pointer value of another pointer

Question

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 ?

simonc · Accepted Answer · 2013-05-28 15:09:17Z

4

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();

edited May 28, 2013 at 15:09

answered May 28, 2013 at 15:03

simonc

42.3k12 gold badges91 silver badges105 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Shafik Yaghmour · Accepted Answer · 2013-05-28 15:04:12Z

2

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 *.

answered May 28, 2013 at 15:04

Shafik Yaghmour

160k44 gold badges466 silver badges778 bronze badges

Comments

Nick · Accepted Answer · 2013-05-28 15:17:08Z

-2

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.

edited May 28, 2013 at 15:17

answered May 28, 2013 at 15:07

Nick

10.6k5 gold badges52 silver badges95 bronze badges

2 Comments

This isn't my real name Over a year ago

This doesn't seem to answer the question. The questioner is asking about getting a pointer back from a called function, but your answer is about not getting a pointer back from the called function.

Nick Over a year ago

Sure, I noticed that. In the way I suggest the (hidden) pointer is send to the called function - opposite to what was asked.

Collectives™ on Stack Overflow

how to assign pointer value of another pointer

3 Answers 3

Comments

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related