correct way to return two dimensional array from a function c

Question

I have tried this but it won't work:

#include <stdio.h>

    int * retArr()
    {
    int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    return a;
    }

    int main()
    {
    int a[3][3] = retArr();
    return 0;
    }

I get these errors:

Error 3 error C2075: 'a' : array initialization needs curly braces
4 IntelliSense: return value type does not match the function type

What am I doing wrong?

Answer 1

A struct is one approach:

struct t_thing { int a[3][3]; };

then just return the struct by value.

Full example:

struct t_thing {
    int a[3][3];
};

struct t_thing retArr() {
    struct t_thing thing = {
        {
            {1, 2, 3},
            {4, 5, 6},
            {7, 8, 9}
        }
    };

    return thing;
}

int main(int argc, const char* argv[]) {
    struct t_thing thing = retArr();
    ...
    return 0;
}

---

The typical problem you face is that int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}}; in your example refers to memory which is reclaimed after the function returns. That means it is not safe for your caller to read (Undefined Behaviour).

Other approaches involve passing the array (which the caller owns) as a parameter to the function, or creating a new allocation (eg using malloc ). The struct is nice because it can eliminate many pitfalls, but it's not ideal for every scenario. You would avoid using a struct by value when the size of the struct is not constant or very large.

Answer 2

#include <stdio.h>

int** retArr()
{
    static int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    return a;
}

int main()
{
    int** a = retArr();
    return 0;
}

You could also specify the returned variable as static. You also must declare the variable a in main as just int** (and not specify the dimensions).

Answer 3

Several issues:

First of all, you cannot initialize an array from a function call; the language definition simply doesn't allow for it. An array of char may be initialized with a string literal, such as

char foo[] = "This is a test";

an array of wchar_t , char16_t , or char32_t may be initialized with a wide string literal, but otherwise the initializer must be a brace-enclosed list of values.

Secondly, you have a type mismatch; except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T " will be converted to an expression of type "pointer to T ", and the value of the expression will be the address of the first element;

In the function retArr , the type of a in the expression is "3-element array of 3-element array of int "; by the rule above, this will be converted to an expression of type "pointer to 3-element array of int ", or int (*)[3] :

int (*retArr())[3]
{
  int a[3][3] = ...;
  return a;
}

but as Brendan points out, once retArr exits a no longer exists; the pointer value that is returned winds up being invalid.

Answer 4

We can solve it by using Heap memory / memory allocating using stdlib :

Allocating memory using stdlib can be done by malloc and calloc.

creating array :

•  int ** arr=( int * * ) malloc ( sizeof ( int * ) * 5 ); 

  is for allocating 5 rows.

•  arr[i]=(int *)malloc(sizeof(int)*5); 

  is for allocating 5 columns in each "i" row.
• Thus we created arr [ 5 ] [ 5 ].

returning array:

•  return arr; 

We just need to send that pointer which is responsible for accessing that array like above.

#include<stdio.h>
#include<stdlib.h>
int **return_arr()
{
    int **arr=(int **)malloc(sizeof(int *)*5);
    int i,j;
    for(i=0;i<5;i++)//checking purpose
    {
        arr[i]=(int *)malloc(sizeof(int)*5);
        for(j=0;j<5;j++)
        {
            arr[i][j]=i*10+j;
        }
    }
    return arr;
}
int main()
{
    int **now;
    now=return_arr();
    int i,j;
    for(i=0;i<5;i++)
    {
        for(j=0;j<5;j++)
        {
            printf("%d ",now[i][j]);
        }
        printf("\n");
    }
    return 0;
}

Answer 5

In the example given by the OP, it is the easiest to put the array on the stack. I didn't test this but it would be like this.

#include stdio.h

void retArr(a[][3]) {
  a[0][0] = 1; a[0][1] = 2; a[0][2] = 3;
  a[1][0] = 4; a[1][1] = 5; a[1][2] = 6;
  a[2][0] = 7; a[2][1] = 8; a[2][2] = 9;
}

main() {
  int a[3][3];
  retArr(a);
}

Yeah, this doesn't "return" the array with the return function, but I would suppose that wasn't the intent of the question. The array, a[][], does get loaded, and the loaded a[][] is available in main after retArr() is called, nothing is overwritten.

Answer 6

For this:

int * retArr() {
    int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    return a;
}

The array has local scope and is destroyed as soon as you exit the function. This means that return a; actually returns a dangling pointer.

To fix this, put the array in global scope, or use malloc() and copy the array into the allocated space.

Answer 7

需要返回int** ，而不是int 。