What is the output of this <code>switch</code> statement in C? Note the statement before any <code>case</code> label. #include<stdio.h> int main() { int i = 1; switch (i) { printf("Hello "); case 1: printf("Hi "); break; case 2: printf("Bye "); break; } return 0; }

Introduction / Context:This question checks knowledge of switch control flow. Statements inside a switch that precede any case label are not “default prelude” statements; control transfers directly to the matching case label, skipping earlier statements.

Given Data / Assumptions:

• i = 1.
• There is a printf("Hello"); placed before the case labels.
• Standard C switch semantics apply: jump to the matching label and execute from there.

Concept / Approach:When switch (i) is executed, the implementation looks for a matching case label (here case 1) and transfers control directly to that label. No statement before that label will run. Therefore, only the statements under case 1 (up to the break) execute.

Step-by-Step Solution:

Verification / Alternative check:Move printf("Hello"); after case 1: and before the printf("Hi");. Then both lines will print; this confirms the role of label placement.

Why Other Options Are Wrong:

Common Pitfalls:Assuming topmost statements run before label dispatch; forgetting that code before the first case is dead unless jumped to.

Final Answer:Hi.