Returning Array from Function

This is a discussion on Returning Array from Function within the C Programming forums, part of the General Programming Boards category; can any one please explain me this code :- Code: #include<stdio.h> #include<conio.h> #define ROW 3 #define COL 4 int main() ...

  1. #1
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    Jul 2011
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    Returning Array from Function

    can any one please explain me this code :-

    Code:
    #include<stdio.h>
    #include<conio.h>
    
    #define ROW 3
    #define COL 4
    
    int main()
    
    {
    	int i,j;
    	int(*c)[ROW][COL];
    	int (*fun3())[ROW][COL];
    	
    
    	c=fun3();
    	printf("Array c[][] in main():\n");
    	for(i=0;i<ROW;i++)
    	{
    	for(j=0;j<COL;j++)
    		printf("%d\n",(*c)[i][j]);
    	printf("\n");
    	}
    	_getch();
    	return 0;
    }
    
    int (*fun3())[ROW][COL]
    {
    	static int c[ROW][COL]={ 
    								6,3,9,1,
    								2,1,5,7,
    								4,1,1,6
    
    							};
    
    int i,j;
    printf("Aray c[][] in fun3():\n");
    for(i=0;i<ROW;i++)
    {
    	for(j=0;j<COL;j++)
    	{
    		printf("%d",c[i][j]);
    	printf("\n");
    	}
    }
    return (int(*)[ROW][COL])c;
    }

    i am not getting
    1.
    Code:
    int (*fun3())[ROW][COL];

    2. working of this :-

    Code:
     for(i=0;i<ROW;i++)
    	{
    	for(j=0;j<COL;j++)
    		printf("%d\n",(*c)[i][j]); 
    3.
    Code:
     return (int(*)[ROW][COL])c;
    please explain and suggest some site/blog/book where i can read on this

  2. #2
    Registered User
    Join Date
    Jun 2005
    Posts
    6,643
    You're probably mixing up types, although I haven't looked closely. A typedef is usually a handy helper to avoid confusion, particularly if you are playing with functions returning pointers to non-trivial things.
    Code:
    #include <stdio.h>
    
    #define ROW 3
    #define COL 4
    
    
    typedef int YourArray[ROW][COL];
    
    int main()
    {
        YourArray *fun3();
        YourArray *c = fun3();
         /* print out array as in your code */
        return 0;
        
    }
    
    YourArray *fun3()
    {
    	static int c[ROW][COL]={ 
    								6,3,9,1,
    								2,1,5,7,
    								4,1,1,6
    
    							};
             return &c;
    }
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

  3. #3
    Banned
    Join Date
    Aug 2010
    Location
    Ontario Canada
    Posts
    9,547
    First C cannot return an array from a function... it doesn't know how. The best you can get is a pointer which may or may not be valid after the function returns... Here's a little test program you can run to see why you should not do this...

    Code:
    #include <stdio.h>
    
    int* MyFunction(int a, int b, int c)
      {  static int array[3];
         array[0] = a;
         array[1] = b;
         array[2] = c;
         return array;  } // return a pointer.
    
    
    int main (void)
      { int *a1, *a2;  // int pointers
    
        printf("calling a1 = MyFunction(10,20,30);\t");
        a1 = MyFunction(10,20,30);
        printf("a1 has %d %d %d\n",a1[0],a1[1],a1[2]);
    
        printf("calling a2 = MyFunction(100,200,300);\t");
        a2 = MyFunction(100,200,300);
        printf("a2 has %d %d %d\n",a2[0],a2[1],a2[2]);
    
        printf("\nLooks good, except...\t"); 
        printf("a1 now has %d %d %d\n",a1[0],a1[1],a1[2]);
    
        getchar();
        return 0; }

  4. #4
    kotin
    Join Date
    Oct 2009
    Posts
    132
    Quote Originally Posted by tarunjain07 View Post
    can any one please explain me this code :-

    Code:
    #include<stdio.h>
    #include<conio.h>
    
    #define ROW 3
    #define COL 4
    
    int main()
    
    {
    	int i,j;
    	int(*c)[ROW][COL];
    	int (*fun3())[ROW][COL];
    	
    
    	c=fun3();
    	printf("Array c[][] in main():\n");
    	for(i=0;i<ROW;i++)
    	{
    	for(j=0;j<COL;j++)
    		printf("%d\n",(*c)[i][j]);
    	printf("\n");
    	}
    	_getch();
    	return 0;
    }
    
    int (*fun3())[ROW][COL]
    {
    	static int c[ROW][COL]={ 
    								6,3,9,1,
    								2,1,5,7,
    								4,1,1,6
    
    							};
    
    int i,j;
    printf("Aray c[][] in fun3():\n");
    for(i=0;i<ROW;i++)
    {
    	for(j=0;j<COL;j++)
    	{
    		printf("%d",c[i][j]);
    	printf("\n");
    	}
    }
    return (int(*)[ROW][COL])c;
    }
    i am not getting
    1.
    Code:
    int (*fun3())[ROW][COL];

    Hi this is function declaration. This meaning function(fun3) returning an pointer to an two dimentional array of width(row) and height(col)

    2. working of this :-

    Code:
     for(i=0;i<ROW;i++)
    	{
    	for(j=0;j<COL;j++)
    		printf("%d\n",(*c)[i][j]); 
    here you are going to print the values in two dimensional array. Actually c is pointer to two dimensional array. so you can get those values by (*c)[i][j].

    3.
    Code:
     return (int(*)[ROW][COL])c;

    Here the function returning the pointer(c) to an two dimensional array. Because we already declared this function prototype in main fucntion.

    please explain and suggest some site/blog/book where i can read on this
    Last edited by nkrao123@gmail.; 09-03-2011 at 09:03 AM.

  5. #5
    Registered User
    Join Date
    Oct 2008
    Location
    TX
    Posts
    2,058
    Quote Originally Posted by tarunjain07 View Post
    i am not getting
    ...
    Code:
     return (int(*)[ROW][COL])c;
    That cryptic cast is pointless, when just returning a pointer to "c" is enough:
    Code:
    return &c;

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