Return an array of char in C and malloc usage

Question

First attempt:

char* loadValues (char* str) {

  char* toReturn[5];

  .. some operations here ..

  return toReturn

}

This will obviously return warnings and will not work properly since the memory location is going to be free-d after the function is finished.

So I thought of using malloc, however, I don't understand how this may work with arrays.

Second attempt:

char* loadValues (char* str) {

  char (*toReturn)[5] = malloc(sizeof *toReturn);

  .. some operations here ..

  return toReturn

}

my toReturn contains strings, for example toReturn[0] may be "Hello"

Any suggestion?

your function is supposed to return pointer to char but your toReturn variable is an array of pointers to char — Andrey Chernukha
– Andrey Chernukha, Commented Jan 10, 2013 at 18:45
possible duplicate of How do I dynamically allocate an array of strings in C? — netcoder
– netcoder, Commented Jan 10, 2013 at 18:49

P.P · Accepted Answer · 2013-01-10 19:19:32Z

4

As far as I understand, you want to return an array of pointers and allocate memory for the pointees of that array. With your current code, you can't return the array of pointers as it's local. You can do it in the following way:

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

char** loadValues () {
  char** toReturn;
  int i;
  toReturn = malloc(5*sizeof(char*));
  for(i=0;i<5;i++)
  {
    toReturn[i] = malloc(25); //Change the size as per your need
    strncpy(toReturn[i], "string",i+1); //Something to copy
  }
  return toReturn;
}

int main()
{
  int i; 
  char **arr = loadValues();
  for(i=0;i<5;i++)
  {
    printf("%s\n", arr[i]);
  }

  for(i=0;i<5;i++)
  {
    free(arr[i]);
  }

  free(arr);
  return 0;  
}

Notice the return type of loadValues and memory allocated for the array is freed in main.

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

void loadValues (char **toReturn) {
  int i;
  for(i=0;i<5;i++)
  {
    toReturn[i] = malloc(25); //Change the size as per your need
    strncpy(toReturn[i], "string",i+1); //Something to copy
  }
}

int main()
{
  int i; 
  char *arr[5];
  loadValues(arr);
  for(i=0;i<5;i++)
  {
    printf("%s\n", arr[i]);
  }

  for(i=0;i<5;i++)
  {
    free(arr[i]);
  }

  return 0;  
}

You should also check if calls to malloc succeeded and handle errors.

edited Jan 10, 2013 at 19:19

answered Jan 10, 2013 at 19:02

P.P

122k20 gold badges193 silver badges251 bronze badges

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

2 Comments

piggyback Over a year ago

at this stage isnt that better to declare the variable char* toReturn[5]; in main and then pass it through loadValues(toReturn) ?

P.P Over a year ago

You can do that as well. Whether you allocate array of pointers or pass it from main is your choice. With a simple modification, you can change the code to pass an array of pointers to loadValues.

user1576605 · Accepted Answer · 2013-01-10 18:49:14Z

1

malloc simply works by allocating a chunk of memory and returning a pointer to the beginning. What might be tricky here is that the array you're making doesn't contain characters, it contains pointers to characters, which means you can't simply say

char* toReturn[5] = malloc(memory);

As far as I can see you'd need to malloc enough memory for five char pointers for toReturn, then go through the elements and malloc however much memory you think you'll need to them. Take extra care you remember to go through the array again and free all the memory when you're done with it, before freeing the array itself.

answered Jan 10, 2013 at 18:49

user1576605

413 bronze badges

Collectives™ on Stack Overflow

Return an array of char in C and malloc usage

2 Answers 2

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related