Const-correctness and arrays: Given a const int array in Turbo C and a function fun(int *), what is the outcome of calling fun(&arr[3]) after printing arr[3]?

Introduction / Context:Const-correctness ensures that objects declared read-only are not modified inadvertently. When a function takes an int *, passing it the address of a const int violates that promise. Even though some legacy compilers accept such code with only a warning, the intent of the language is to forbid discarding const implicitly. This question examines the compile-time consequence of attempting to modify a const array element through a non-const pointer parameter.

Given Data / Assumptions:

• const int arr[5] = {1,2,3,4,5};
• int fun(int *f) { *f = 10; return 0; }
• Call site: fun(&arr[3]); after printing arr[3].

Concept / Approach:

• The type of &arr[3] is const int *. Converting to int * would allow writing through the pointer to a const object, which is disallowed.
• Modern compilers issue an error; older ones might warn but still compile, leading to undefined behavior at runtime if a write occurs.
• Therefore, the correct, standard-conforming outcome is a type error at the call.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

• ”Before…4 After…10”: implies the illegal write succeeded.
• ”Invalid parameter”: vague, not the standard type mismatch description.
• ”Before…4 After…4”: would require that the write be ignored; the core issue is a compile-time type error.

Common Pitfalls:

• Assuming that a warning is harmless; writing through a pointer to const invokes undefined behavior.
• Confusing array decays and pointer types; the const-qualification applies to the pointed-to object.

Final Answer: