C programming: what will be the output of this program on a standards-compliant compiler? #include int main() { const c = -11; const int d = 34; printf("%d, %d ", c, d); return 0; }

Introduction / Context:This question checks your understanding of declarations, type qualifiers (const), and compilation rules in C. While many compilers offer extensions, a strictly conforming C compiler requires that every object declaration include a type. Printing behavior only matters if the code compiles, so the key concept is whether the declaration of c is valid C.

Given Data / Assumptions:

• c is declared as const c = -11; (no explicit type).
• d is declared as const int d = 34; (valid).
• printf uses the %d format for both values.
• Assume a standard-conforming compiler (ISO C), not compiler-specific extensions.

Concept / Approach:

In ISO C, const is a type qualifier that must qualify a complete type such as int, double, etc. A declaration like const c = -11; omits the base type, making the declaration ill-formed and causing a compile-time error. Some historical or non-conforming modes might treat it as int, but that is not guaranteed by the C standard.

Step-by-Step Solution:

Verification / Alternative check:

Try compiling with a standards-conforming mode (for example, enable ISO C warnings). The compiler will reject the declaration of c. Changing it to const int c = -11; makes the program valid and then the output would be -11, 34.

Why Other Options Are Wrong:

-11, 34: Would only be correct if c had a valid type, e.g., const int.

11, 34: The sign would not flip; there is no operation that negates -11.

None of these: The precise correct choice is “Error”.

Common Pitfalls:

• Assuming const implies int by default—this is not guaranteed by the standard.
• Testing on permissive compilers and generalizing the behavior as portable.

Final Answer:

Error