[Solved] Consider the following C code:int x = 1; void outer() { int x = 2; { int x = 3; printf("%d ", x); } printf("%d ", x); } int main() { outer();... }What is the output when this program runs? | C

C - Variables and Data Types

Consider the following C code:

int x = 1;
void outer() {
    int x = 2;
    {
        int x = 3;
        printf("%d ", x);
    }
    printf("%d ", x);
}
int main() {
    outer();
    printf("%d", x);
    return 0;
}

What is the output when this program runs?

A1 2 3

B3 2 1

C3 3 3

D2 3 1

Step-by-Step Solution

Solution:

Step 1: Analyze innermost block variable
The innermost block declares x = 3, so first printf prints 3.
Step 2: Analyze outer block variable
After inner block ends, the x = 3 goes out of scope, so next printf prints outer x = 2.
Step 3: Analyze global variable in main()
Finally, main prints global x = 1.
Final Answer:
3 2 1 -> Option B
Quick Check:
Inner blocks hide outer variables [OK]

Quick Trick: Inner block variables override outer ones [OK]

Common Mistakes:

Ignoring block-level scope
Assuming all x print same value
Confusing order of prints

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More C Quizzes

Consider the following C code:

Step 1: Analyze innermost block variable

Step 2: Analyze outer block variable

Step 3: Analyze global variable in main()

Final Answer:

Quick Check:

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