Type conversions with nested calls and assignment to a float: what prints? #include<stdio.h> int fun(int); int main() { float k = 3; fun(k = fun(fun(k))); printf("%f ", k); return 0; } int fun(int i) { i++; return i; }

Introduction / Context:
This question examines integer-to-float conversions, nested calls, and assignment expression values. The helper function increments an integer and returns the result. The key point is which value actually gets stored into the float variable k.

Given Data / Assumptions:

• fun(int) returns its argument + 1.
• k is a float initially 3, but all fun calls take and return int.
• Statement: fun(k = fun(fun(k))); followed by printf("%f", k);

Concept / Approach:
Work from the inside out. Conversions occur at call boundaries. Assignment is an expression whose value is the value assigned. The final fun(...) call’s result is not assigned to k unless explicitly stored.

Step-by-Step Solution:
Inner: fun(k) → argument 3 converted to int 3 → returns 4.Next: fun(4) → returns 5.Assignment: k = 5 (k becomes 5.0; the expression value is 5).Outer call: fun(5) → returns 6, but the result is ignored.Printed value is the stored k = 5.000000.

Verification / Alternative check:
If the code had been k = fun(fun(fun(k))); then k would be 6. The difference is whether the outermost return value is assigned back to k.

Why Other Options Are Wrong:
“3.000000/4.000000” ignore the nested increments and the assignment. “Garbage value” does not apply; behavior is defined. “6.000000” would require assigning the outer call’s result to k.

Common Pitfalls:
Confusing what gets assigned vs. what gets computed; overlooking implicit conversions at call sites.

Final Answer:
5.000000