Partial structure initialization in C: unnamed members default to zero. What does this print? #include int main() { struct emp { char name[20]; int age; float sal; }; struct emp e = {"Tiger"}; // only name given; others default-initialized printf("%d, %f ", e.age, e.sal); return 0; }

Introduction / Context:Structures in C can be partially initialized with aggregate initializers. Understanding what happens to unspecified members avoids undefined behavior and misinterpretation of output.

Given Data / Assumptions:

• A struct has three members: char name[20]; int age; float sal;
• Initializer provides only the first member: {"Tiger"}.
• Printing the numeric members age and sal.

Concept / Approach:In an aggregate initializer, any members not explicitly initialized are initialized as if by zero. This rule applies regardless of storage duration when an initializer is present. Therefore, age becomes 0 and sal becomes 0.0.

Step-by-Step Solution:Assign name = "Tiger".Omitted members age and sal → set to 0 and 0.0 respectively.Print “0, 0.000000”.

Verification / Alternative check:Explicitly writing struct emp e = {"Tiger", 0, 0.0f}; yields identical behavior; both are well-defined.

Why Other Options Are Wrong:B: Not garbage—there is an initializer. C: No syntax error. D/E: The behavior is specified by the C standard and is not implementation dependent in this context.

Common Pitfalls:Confusing uninitialized automatic structs (indeterminate) with partially initialized ones (rest zero). Forgetting the difference between designated and positional initializers.

Final Answer:0, 0.000000