Accessing elements in a 2D array in C-CodePudding

I am writing a program that passes a 2D char array into a function for printing. However, when I try to access elements in the 2D array for printing, I can't. This problem does not occur when I place the printing loops and statements in the main() function, only if the array is passed. Can someone help explain?

void disp_array(char* array)
{
    for (int i = 0; i < SIZE; i  )
    {
        for (int j = 0; j < SIZE; j  )
        {
            printf("%c", array[i][j]);
        }
        printf("\n");
    }
}

Attached is my code, the j in the printf statement is highlighted with an error - E0142: "expression must have pointer-to-object type but it has type int"

CodePudding user response：

The function declaration is wrong.

As the parameter has the type char * then the expressions array[i] yields a scalar object of the type char to which you may not apply the subscript operator as you are trying to do array[i][j]

The function should be declared like

void disp_array(char array[][SIZE], int n )
{
    for (int i = 0; i < n; i  )
    {
        for (int j = 0; j < SIZE; j  )
        {
            printf("%c", array[i][j]);
        }
        printf("\n");
    }
}

or like

void disp_array(char (*array)[SIZE], int n )
{
    for (int i = 0; i < n; i  )
    {
        for (int j = 0; j < SIZE; j  )
        {
            printf("%c", array[i][j]);
        }
        printf("\n");
    }
}

And along with the array you need to pass the number of rows in the array. For example

disp_array( array, SIZE );

If the array contains strings then the function definition will look like

void disp_array(char (*array)[SIZE], int n )
{
    for (int i = 0; i < n; i  )
    {
        puts( array[i] );
    }
}

CodePudding user response：

Although a 1D array in C "decays" to a pointer when used as a function argument, this is not true for a 2D array – which would decay to an array of pointers.

When you have a function that takes a (fixed) array as an argument (whatever its dimensions), just declare that argument as an array – and let the compiler sort out any necessary 'conversion' to pointers.

In your case (I have assumed a value of 4 for SIZE) just declare the argument for what it is: char array[SIZE][SIZE]:

#include <stdio.h>

#define SIZE 4

void disp_array(char array[SIZE][SIZE])
{
    for (int i = 0; i < SIZE; i  ) {
        for (int j = 0; j < SIZE; j  ) {
            printf("%c", array[i][j]);
        }
        printf("\n");
    }
}

int main()
{
    char test[SIZE][SIZE] = {
        {'a','b','c','d'},
        {'e','f','g','h'},
        {'i','j','k','l'},
        {'m','n','o','p'}
    };
    disp_array(test);
    return 0;
}

In this case, if you have an IDE that will let you see what the actual signature of disp_array is, it will show (something like) void disp_array(char (*array)[4]).