in this code:

#include <stdio.h>

int main(void) 
{
    int a[2][3] = {{1,2},{4,5,6}};

    printf("%d\n",a[0][2]);
    return 0;
}

The output is 0 - since the array wasn't initialized, is this output a result of some undefined behaviour?

CodePudding user response：

From the C Standard (6.7.9 Initialization)

19 The initialization shall occur in initializer list order, each initializer provided for a particular subobject overriding any previously listed initializer for the same subobject;151) all subobjects that are not initialized explicitly shall be initialized implicitly the same as objects that have static storage duration.

and

10 If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static or thread storage duration is not initialized explicitly, then:

— if it has arithmetic type, it is initialized to (positive or unsigned) zero;

So in fact this declaration

int a[2][3] = {{1,2},{4,5,6}};

is equivalent to

int a[2][3] = {{1,2, 0},{4,5,6}};

CodePudding user response：

The array was initialised. You’re defining a two-dimensional array that consists of two arrays of three integers each. For the first of these two you only give two values, and by default the missing one is initialised as zero, so your full array is {{1, 2, 0}, {4, 5, 6}. a[0][2] will give you that zero.