how to return a char array from a function in C

Question

I want to return a character array from a function. Then I want to print it in main. how can I get the character array back in main function?

#include<stdio.h>
#include<string.h>
int main()
{
    int i=0,j=2;
    char s[]="String";
    char *test;

    test=substring(i,j,*s);   
    printf("%s",test);
    return 0;
}


char *substring(int i,int j,char *ch)
{
    int m,n,k=0; 
    char *ch1;
    ch1=(char*)malloc((j-i+1)*1);
    n=j-i+1;

    while(k<n)
    {   
        ch1[k]=ch[i];
        i++;k++;
    }   

    return (char *)ch1;
}

Please tell me what am I doing wrong?

Your compiler should have told you that test=substring(i,j,*s); you're dereferencing s when you shouldn't. — Daniel Fischer
– Daniel Fischer, Commented May 8, 2013 at 14:51
i am getting an error message : conflicting types for ‘substring’ — sayan
– sayan, Commented May 8, 2013 at 14:52
Unrelated to errors, but I believe both your (char*) casts are unnecessary and multiplication by 1 is pointless - (j-i+1)*1 should be simplified to j-i+1. — Bernhard Barker
– Bernhard Barker, Commented May 8, 2013 at 14:54
In addition to the other comments, you need to add a null terminator to ch1 at the end of the loop in substring — simonc
– simonc, Commented May 8, 2013 at 14:55

Coffee_lover · Accepted Answer · 2018-06-24 20:04:58Z

10

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char *substring(int i,int j,char *ch)
{
    int n,k=0;
    char *ch1;
    ch1=(char*)malloc((j-i+1)*1);
    n=j-i+1;

    while(k<n)
    {
        ch1[k]=ch[i];
        i++;k++;
    }

    return (char *)ch1;
}

int main()
{
    int i=0,j=2;
    char s[]="String";
    char *test;

    test=substring(i,j,s);
    printf("%s",test);
    free(test); //free the test 
    return 0;
}

This will compile fine without any warning

#include stdlib.h
pass test=substring(i,j,s);
remove m as it is unused
either declare char substring(int i,int j,char *ch) or define it before main

edited Jun 24, 2018 at 20:04

answered May 8, 2013 at 15:10

Coffee_lover

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2 Comments

Coffee_lover Over a year ago

it will compile fine without any warning //1.>include stdlib.h //2.>pass test=substring(i,j,s); //3.>remove m as it is unused //4.>either declare char substring(int i,int j,char *ch) or define it before main

Patryk Over a year ago

Is it only me who sees the memory leak on test? (after 2,5 years ;) )

ShinTakezou · Accepted Answer · 2013-05-08 15:04:53Z

Lazy notes in comments.

#include <stdio.h>
// for malloc
#include <stdlib.h>

// you need the prototype
char *substring(int i,int j,char *ch);


int main(void /* std compliance */)
{
  int i=0,j=2;
  char s[]="String";
  char *test;
  // s points to the first char, S
  // *s "is" the first char, S
  test=substring(i,j,s); // so s only is ok
  // if test == NULL, failed, give up
  printf("%s",test);
  free(test); // you should free it
  return 0;
}


char *substring(int i,int j,char *ch)
{
  int k=0;
  // avoid calc same things several time
  int n = j-i+1; 
  char *ch1;
  // you can omit casting - and sizeof(char) := 1
  ch1=malloc(n*sizeof(char));
  // if (!ch1) error...; return NULL;

  // any kind of check missing:
  // are i, j ok? 
  // is n > 0... ch[i] is "inside" the string?...
  while(k<n)
    {   
      ch1[k]=ch[i];
      i++;k++;
    }   

  return ch1;
}

user1508519 · Accepted Answer · 2013-05-08 14:53:00Z

4

Daniel is right: http://ideone.com/kgbo1C#view_edit_box

Change

test=substring(i,j,*s);

to

test=substring(i,j,s);

Also, you need to forward declare substring:

char *substring(int i,int j,char *ch);

int main // ...

answered May 8, 2013 at 14:53

user1508519

1 Comment

sayan Over a year ago

im getting right result for integer type array. But when im trying the same with a char array im getting the error only

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