In C programming, what is the result when you compile the following code that uses variable expressions in switch case labels? #include <stdio.h> int main() { int a = 1, b = 1; switch (a) { case a * b: printf("yes "); case a - b: printf("non"); break; } return 0; }

Introduction / Context:
This question examines your understanding of the rules for switch statements in C, specifically the requirement that case labels must be integer constant expressions. It also checks whether you can recognize invalid uses of variables inside case labels that lead to compilation errors rather than runtime outputs.

Given Data / Assumptions:

• a and b are integer variables, both initialized to 1.

• The switch expression is switch (a).

• The case labels are written as case a * b: and case a - b:.

• The code attempts to print "yes " and "non" inside these cases.

• We assume a standard C compiler that follows the language rule for case labels.

Concept / Approach:
In C, each case label in a switch statement must be an integer constant expression that can be evaluated at compile time. That means it can use only constants, enumeration constants and the result of operators applied to them, but not ordinary variables whose values are only known at runtime. Using a variable such as a or b directly in the case expression violates this rule and forces the compiler to issue a diagnostic, resulting in a compile time error.

Step-by-Step Solution:
Step 1: Note that the switch expression itself, switch (a), can use a variable; that is allowed. Step 2: Examine the first case label: case a * b:. Here a and b are ordinary int variables, not constants or enums. Step 3: Because a * b depends on variable values, it is not an integer constant expression, so it is invalid as a case label. Step 4: The same reasoning applies to case a - b:, which also uses variables and is not a constant expression. Step 5: A conforming compiler must reject this code with a diagnostic, so the program does not successfully compile and therefore produces no runtime output.

Verification / Alternative check:
As a quick test, you can rewrite the code using const or #define: for example, define C1 1 and C2 1 and then write case C1 * C2:. This version uses constants and will compile successfully. The fact that the original version using variables does not compile confirms that the issue is specifically about the requirement for case labels to be constant expressions, not about the switch mechanics themselves.

Why Other Options Are Wrong:
Option A ("yes") assumes that case a * b: is valid and that execution falls into that case, which is not possible because the code never compiles. Option B ("non") similarly assumes valid compilation and execution reaching case a - b:, which also cannot happen. Option D ("yes non") assumes both prints occur due to fall through, which again presumes successful compilation. All three options ignore the fundamental rule about case labels needing to be constant, making them incorrect.

Common Pitfalls:
A frequent mistake is to think that any integer expression is allowed in a case label, regardless of whether it is constant or uses variables. Another pitfall is to focus solely on the runtime behavior of the switch and ignore the compile time constraints. Remember that switch case values must be fixed at compile time so the compiler can build an efficient jump table or decision tree. Always check whether an expression uses variables or only constants before using it in a case label.

Final Answer:
Because the case labels use variables and are not constant expressions, the code causes a compile time error.