How to malloc inside a function and return pointer in C?

Question

Below is some psudo, but I'm trying to accomplish this. The problem is as written, it returns a blank pointer.

int testFunction(char *t) {
    int size = 100;
    t = malloc(100 + 1);
    t = <do a bunch of stuff to assign a value>;
    return size;
}

int runIt() {
    char *str = 0;
    int str_size = 0;
    str_size = testFunction(str);
    <at this point, str is blank and unmodified, what's wrong?>
    free(str);
    return 0;
}

This works fine if I have a predefined size, such as char str[100] = "" and I don't try to malloc or free memory afterwords. I need to be able to make the size dynamic though.

I've also tried this, but seem to run into a corrupt pointer somehow.

int testFunction(char **t) {
    int size = 100;
    t = malloc(100 + 1);
    t = <do a bunch of stuff to assign a value>;
    return size;
}

int runIt() {
    char *str = 0;
    int str_size = 0;
    str_size = testFunction(&str);
    <at this point, str is blank and unmodified, what's wrong?>
    free(str);
    return 0;
}

Thanks!

John Carter · Accepted Answer · 2012-10-09 00:58:46Z

6

You're nearly there with the second example, but change

int testFunction(char **t) {
  ...
  t = malloc(100 + 1);

To

int testFunction(char **t) {
  ...
  *t = malloc(100 + 1);

The point being that you're passing in a char**, a pointer to a pointer, so you want to assign the malloc to what that points at (a pointer).

answered Oct 9, 2012 at 0:58

John Carter

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Comments

Mark Tolonen · Accepted Answer · 2012-10-09 12:10:10Z

6

Your test function is just a bit backward. Size should be an input. The allocated pointer should be the output:

char* testFunction(int size) {
    char* p = malloc(size);
    <do a bunch of stuff to assign a value>;
    return p;
}

int runIt() {
    char *str = 0;
    int str_size = 100;
    str = testFunction(str_size);
    <do something>
    free(str);
    return 0;
}

edit

Per comment, making size an output too.

char* testFunction(int *size) {
    *size = <compute size>;
    char* p = malloc(size);
    <do a bunch of stuff to assign a value>;
    return p;
}

int runIt() {
    char *str = 0;
    int str_size;
    str = testFunction(&str_size);
    <do something>
    free(str);
    return 0;
}

edited Oct 9, 2012 at 12:10

answered Oct 9, 2012 at 0:57

Mark Tolonen

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3 Comments

Fmstrat Over a year ago

This would have been a good option except in my code the variable "size" must be determined in testFunction. The marked answer is a solution specific to my issue. Thanks.

Yash Jain Over a year ago

You completely changed the paramters. Incorrect solution

Mark Tolonen Over a year ago

@YashJain I changed the parameters because the question said he needed to make the size dynamic. The size needs to be passed in as an input or returned as an output. Per the OP's comment the 2nd option would be viable. The accepted solution doesn't allow a dynamic size (it's fixed). There's nothing wrong with the solution.

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