What will be the output of the following C code? #include <stdio.h> void main() { int x = 0; int *ptr = &x; printf("%d\n", *ptr); }

The output of the given C code will be "0".

Here's the step by step explanation:

1. The integer variable x is declared and initialized with the value 0.
2. A pointer variable *ptr of type integer is declared and it is assigned the address of x using the & operator.
3. The printf function is used to print the value at the address stored in the pointer ptr. The * operator is used to dereference the pointer, i.e., to get the value stored at the address it is pointing to.
4. Since ptr is pointing to x, and x is 0, the printf function prints 0.
5. So, the output of the program is 0.