Explain me this output

Question

Int arr[2][3] = {2,3,4,5,6,7};
Printf("%d",(*(*arr+1))));

Answer 1

The output should be a compiler error. There are too many closing parentheses on the printf statement. Also int and printf should not be capitalized, but I assume your keyboard added capitalization.

Here is the corrected code. I added braces to clarify the row groupings:

int arr[2][3] = {{2,3,4},{5,6,7}};
printf("%d",(*(*arr+1)));

Since arr is a two-dimensional array, think of it as a pointer to row pointers. To reach the value of an element, it requires two dereferences. Here is how it occurs above:

*arr is a pointer to the first row {2,3,4}
(*arr)+1 shifts the row pointer to the second element of the row
*((*arr)+1) returns the contents, 3

I added parentheses to clarify the order of operations. It is the same as *(*arr+1).

Answer 2

Output is 3.
Poienter *arr indicate arr [0]=2;
*arr+1 indicate arr [1]=3.
Same thing do the *(*arr+1)

Answer 3

your code indicating error. because acc to statement shows two dimensional array arr[2][3]
two rows and three column but arr[2][3] = {{2,3,4},{5,6,7}}; you code should be this.