Binding a const to a non-const value: After int y = 128; const int x = y; what is printed by printf("%d ", x)?

Introduction / Context:The const qualifier stops further mutation through the declared name, but it does not alter the stored value at initialization. Assigning or initializing a const from an ordinary variable is routine and results in an independent read-only object that holds the copied value at that moment. This question reinforces that constness affects mutability, not representable values or printing behavior.

Given Data / Assumptions:

• int y = 128;
• const int x = y;
• We print x using %d.

Concept / Approach:

• Initialization copies the value of y into x; x is then read-only.
• There is no type mismatch: both are of type int; const is a qualifier on x's object, not a different base type.
• Therefore, printing x yields 128.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• Garbage value/0: contradicts straightforward initialization.
• Error: no rule forbids initializing a const int from a non-const int.
• Unspecified behavior: there is no unspecified behavior in this sequence.

Common Pitfalls:

• Confusing const int x = y; with binding a reference (a C++ concept). In C, it is simply value initialization.
• Thinking that const implies compile-time constness; here it is a runtime copy but a read-only one thereafter.

Final Answer: