Why it is not printing 2d array value when access directly

Question

#include<stdio.h>
int main(){
int i,j;
int ans[5][5];
ans[0][0] = 10;
printf(" why this %d \n\n",ans[0,0] );
}

and output is some garbage value i tried this in codeblocks

Answer 1

You declared a two-dimensional array

int ans[5][5];

The element type of the array is int[5]. So for example the expression ans[0] gives such an array. Arrays used in expressions (with rare exceptions) are converted to pointers to their first elements. So the expression ans[0] having the type int[5] is implicitly converted to pointer of the type int * to the element ans[0][0]

In this expression

ans[0,0]

in the square brackets there is used an expression with the comma operator 0,0. As the first operand of the expression with the comma operator does not have a side effect then the above expression is equivalent to ans[0].

That means that in this call

printf(" why this %d \n\n",ans[0,0] );

that is equivalent to

printf(" why this %d \n\n",ans[0] );

you are trying to output a pointer using the conversion specifier %d designed to output objects of the type int that results in undefined behavior.

It seems you mean

printf(" why this %d \n\n",ans[0][0] );

or

printf(" why this %d \n\n", **ans );

or

printf(" why this %d \n\n", *ans[0] );

or

printf(" why this %d \n\n",  ( *ans )[0] );

As for the comma operator then just consider the following code snippet

int i = 0;
ans[0][0] = ( ++i, ++i, ++i, ++i, ++i, ++i, ++i, ++i, ++i, ++i );

the result of which is equivalent to the following code snippet

int i = 10;
ans[0][0] = i;