returning arrays from function using pointer_segmentation fault

This is a discussion on returning arrays from function using pointer_segmentation fault within the C++ Programming forums, part of the General Programming Boards category; Code: #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() ...

  1. #1
    Registered User
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    Jan 2011
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    4

    Smile returning arrays from function using pointer_segmentation fault

    Code:
    #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

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    Your indentation suggests that your for loop on i has more than one statement in it; however, the code does not agree. Use braces.

  3. #3
    -bleh-
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    The display loop is incorrect btw you need to free the array.
    "All that we see or seem
    Is but a dream within a dream." - Poe

  4. #4
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    Smile solved... thanks

    Code:
    #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...

  5. #5
    Registered User
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    Jan 2011
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    4

    Smile passing array to function using pointers

    Code:
    #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
    Last edited by vineeshvs; 01-14-2011 at 10:42 PM.

  6. #6
    -bleh-
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    are you kidding? I thought you fixed it in post #4. Still need to free the arrays.
    "All that we see or seem
    Is but a dream within a dream." - Poe

  7. #7
    Registered User whiteflags's Avatar
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    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.

  8. #8
    -bleh-
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    somewhere in this universe
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    her'es your problem
    Code:
     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
    "All that we see or seem
    Is but a dream within a dream." - Poe

  9. #9
    and the hat of int overfl Salem's Avatar
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    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

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