Question

#include
int main(int argv, char*arg[])
{
int x=280;
char* ptr_p=(char*) &x;
printf("%d\n", *ptr_p);
return 0;
}

Answer 1

The output of the code:

output:

24

Explanation:

• In the given code an integer variable "x" is declared, which initializes a value, that is "280".
• In the next step, another char pointer variable "ptr_p" is defined, which holds the address of variable x.
• At the last, the print method is used that prints variable x address value that is "24".

Learn more:

• Output of the code: https://brainly.in/question/5321421