returning arrays from function using pointer_segmentation fault

[vineeshvs]

#include<stdio.h>
#include<stdlib.h>
int **array();
 main()
{
  int **result=array(),i,j;



  //display result
for(i=0;i<2;i++)
  printf("\n");
	  for(j=0;j<2;j++)
	    printf("%d \t",result[i][j]);

}




//function
int **array()
{
  int nrows=2,ncolumns=2,i,j;


  // printf("hi");

  //memory allocation
  int **array;
	array = malloc(nrows * sizeof(int *));
	if(array == NULL)
		{
		printf("out of memory\n");
		return 0;
		}

	//printf("hi");
	for(i = 0; i < nrows; i++)
	  	{
	  	array[i] = malloc(ncolumns * sizeof(int));
			if(array[i] == NULL)
			{
			printf("out of memory\n");
			return 0;
			}
		}

 


	//create array
	for(i=0;i<2;i++)
	  for(j=0;j<2;j++)
	    array[i][j]=i+j;




	//returning pointer
	return array;
printf("hi");

}

it compiled successfully. but shows segmentation fault on running. please help to find error

[tabstop]

Your indentation suggests that your for loop on i has more than one statement in it; however, the code does not agree. Use braces.

[nimitzhunter]

The display loop is incorrect btw you need to free the array.

[vineeshvs]

#include<stdio.h>
#include<stdlib.h>
int **array();
 main()
{
  int **result=array(),i,j;



  //display result
  for(i=0;i<2;i++)                             //CORRECTED HERE. one extra printf was here
	  for(j=0;j<2;j++)
	    printf("%d \t",result[i][j]);

}




//function
int **array()
{
  int nrows=2,ncolumns=2,i,j;


  // printf("hi");

  //memory allocation
  int **array;
	array = malloc(nrows * sizeof(int *));
	if(array == NULL)
		{
		printf("out of memory\n");
		return 0;
		}

	//printf("hi");
	for(i = 0; i < nrows; i++)
	  	{
	  	array[i] = malloc(ncolumns * sizeof(int));
			if(array[i] == NULL)
			{
			printf("out of memory\n");
			return 0;
			}
		}

 


	//create array
	for(i=0;i<2;i++)
          for(j=0;j<2;j++)
	    array[i][j]=i+j+2;

	    
	



	//returning pointer
	return array;
printf("hi");

}

thank you. apologize for my negligence. this site is very active, like it...

[vineeshvs]

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


void function(int **x);


main()
{
  int nrows=2,ncolumns=2,i,j;


  //memory allocation for x
  int **x=malloc(nrows*sizeof(int*));
  if(x==NULL)
    {
      printf("out of memory\n");
      return 0;
    }
  for(i=0;i<nrows;i++)
    {
      x[i]=malloc(ncolumns*sizeof(int));
      if(x[i]=NULL)
	{
	  printf("out of memory\n");
	  return 0;
	}
    }


  printf("code passed me");//checking


  //define x
  for(i=0;i<2;i++)	  
    for(j=0;j<2;j++)	    
      x[i][j]=i+j+2;
 




  //call function
  function(x); 

}







//function_definition
function(int **x)

{
  int nrows=2,ncolumns=2,i,j,y[2][2];

		  
  for(i=0;i<2;i++)
    for(j=0;j<2;j++)
      y[i][j]=x[i][j]+1;



  //display y
  for(i=0;i<2;i++)
    for(j=0;j<2;j++)
      printf("%d",y[i][j]);		 
}

segmentation fault comes.. help please

[nimitzhunter]

are you kidding? I thought you fixed it in post #4. Still need to free the arrays.

[whiteflags]

If by "segmentation fault comes" you mean it doesn't compile, then yes, I see where things went wrong. Two mistakes that I can point out include the way function() and main() were both written. For function(), since the prototype says

void function(int **x);

then the definition should also say

void function(int **x)

For main(), the standard expects one of the following to be main's signature in C++

int main()
int main(int argc, char *argv[])

or some variation thereof using the const keyword. But the main thing is that main() returns int.

As for an actual segmentation fault, I don't see one.

[nimitzhunter]

her'es your problem

 if(x[i]=NULL) // use "==" to compare.

That's is assignment. You just made all your array NULL. That's why you can't access x[i][j], thus segfault

[Salem]

Still cross-posting
returning arrays from function using pointer_segmentation fault - C