如何获取多维数组的输入并使用指针在C中打印数组？

Question

I am trying to take integer data and store it into a dimensional array but i am unable to do it. Somebody help me please..

I tried using *(*(arr+i) + j) where arr is a pointer to the 2-D array , i and j are the loop variables, I get an error

error: invalid type argument of unary '*' (have 'int') scanf("%d", ( (arr+i) + j));

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

int main(){
        int n,*arr,i,j,k;
        scanf("%d",&n);
        arr = malloc(n*n*sizeof(int));
        memset(arr,0,n*n*sizeof(int));
        for(i=0;i<n;i++){
                for(j=0;j<n;j++){
                        scanf("%d", *(*(arr+i) + j));
                }
        }
        for(i=0;i<n;i++){
                for(j=0;j<n;j++){
                        printf("%d ", *(*(arr+i) + j);
                }
                printf("\n");
        }

}

My input was:

3
11 2 4
4 5 6
10 8 -12

Answer 1

int* arr ... arr = malloc(n*n*sizeof(int)); gives you a "mangled" 2D array - it's actually a 1D array. Meaning you'll have to access it as arr[i*n + j] .

Mangled arrays are mostly a thing of the past though. With modern standard C, you can replace the whole code with this:

int (*arr)[n] = malloc( sizeof(int[n][n]) );
...
for(size_t i=0; i<n; i++)
  for(size_t j=0; j<n; j++)
    scanf("%d", &arr[i][j]);
...
free(arr);

Also note, if you need to zero-initialize the whole array to zero, you are better off using calloc since it does just that.

Answer 2

The type of the variable arr is int * .

int n,*arr,i,j,k;
      ^^^^

So there is no a two-dimensional array in your program.

So for example this expression

*(arr+i) + j

has type int . And this expression

*(*(arr+i) + j)

tries to dereference an object of the type int that is an object that is not a pointer.

If your compiler supports variable length arrays then the program can look like

#include <stdio.h>

int main(void) 
{
    size_t n;

    scanf( "%zu", &n );

    int a[n][n];

    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < n; j++ )
        {
            scanf( "%d", *( a + i ) + j );
        }
    }

    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < n; j++ )
        {
            printf( "%3d ", *( *( a + i ) + j ) );
        }
        putchar( '\n' );
    }

    return 0;
}

Its output is

 11   2   4 
  4   5   6 
 10   8 -12 

Otherwise another approach is dynamically to allocate a one-dimensional array of pointers and then correspondingly one-dimensional arrays of integers.

For example

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

int main(void) 
{
    size_t n;

    scanf( "%zu", &n );

    int **a = malloc( n * sizeof( int * ) );

    for ( size_t i = 0; i < n; i++ )
    {
        *( a + i ) = malloc( n * sizeof( int ) );
    }

    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < n; j++ )
        {
            scanf( "%d", *( a + i ) + j );
        }
    }

    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < n; j++ )
        {
            printf( "%3d ", *( *( a + i ) + j ) );
        }
        putchar( '\n' );
    }

    for ( size_t i = 0; i < n; i++ )
    {
        free( *( a + i ) );
    }

    free( a );

    return 0;
}

The program output will be the same as shown above.