[Solved] What will be the output of this C program?int x = 1; void test() { x = 5; int x = 10; printf("%d ", x); } int main() { test(); printf("%d", x); return 0; } | C

C - Variables and Data Types

What will be the output of this C program?

int x = 1;
void test() {
    x = 5;
    int x = 10;
    printf("%d ", x);
}
int main() {
    test();
    printf("%d", x);
    return 0;
}

A5 1

B10 5

C5 10

D10 1

Step-by-Step Solution

Solution:

Step 1: Understand variable shadowing and assignment
Inside test(), 'int x = 10;' declares a new local x, shadowing global x. The assignment 'x = 5;' before local declaration affects global x.
Step 2: Trace output values
printf inside test() prints local x = 10. After test(), global x was set to 5, so main prints 5.
Final Answer:
10 5 -> Option D
Quick Check:
Local shadows after assignment; global x changed = 10 5 [OK]

Quick Trick: Local variables declared after assignment don't affect earlier assignments [OK]

Common Mistakes:

Master "Variables and Data Types" in C

9 interactive learning modes - each teaches the same concept differently

More C Quizzes

What will be the output of this C program?