What is the output of the following code?int arr[5] = {1, 2, 3};printf("%d\n", arr[4]);

Break Down the Problem

1. Identify the code segment provided.
2. Understand how arrays work in C programming, especially when they are partially initialized.
3. Check what happens when accessing an out-of-bounds index in an array.

Relevant Concepts

1. In C, an array declared with size n can hold n elements.
2. Uninitialized elements in an array are initialized to 0 if the array has static storage duration (e.g., global arrays or static arrays).
3. Accessing an index outside the bounds of the array can lead to undefined behavior.

Analysis and Detail

1. The array arr is declared with size 5 and partially initialized with the values {1, 2, 3}. The remaining two elements will be initialized to 0 since the array has static storage duration by default.

   • arr[0] = 1
   • arr[1] = 2
   • arr[2] = 3
   • arr[3] = 0 (default initialization)
   • arr[4] = 0 (default initialization)

2. The line of code printf("%d\n", arr[4]); attempts to print the value at the index 4 of the array arr, which corresponds to the fifth element of the array.

Verify and Summarize

1. Since arr[4] is initialized to 0, the printf statement will output 0.

Final Answer

The output of the given code will be:

0