Chained relational operators in C: how is this expression evaluated? #include<stdio.h> int main() { int x = 10, y = 20, z = 5, i; i = x < y < z; // left-to-right evaluation with boolean 0/1 printf("%d ", i); return 0; }

Introduction / Context:
C does not support mathematical chained comparisons like some languages. Instead, relational operators are evaluated left-to-right, and each comparison yields either 0 (false) or 1 (true). Understanding this prevents logic mistakes.

Given Data / Assumptions:

• x = 10, y = 20, z = 5
• Expression: x < y < z
• C relational operators yield int 0 or 1.

Concept / Approach:
Evaluation: x < y is 10 < 20 → 1. Then we evaluate 1 < z which is 1 < 5 → 1. Therefore, the final result assigned to i is 1.

Step-by-Step Solution:
Compute x < y → 1.Then evaluate (1) < z → 1 < 5 → 1.Assign i = 1 and print it.

Verification / Alternative check:
To express a mathematical chain correctly, use logical AND: (x < y) && (y < z). With z=5 that expression would be true && false → 0, differing from the chained form.

Why Other Options Are Wrong:
A/D/E mischaracterize the deterministic result. C: It is valid C, not a syntax error.

Common Pitfalls:
Assuming C treats chained comparisons like mathematics; forgetting that intermediate boolean becomes 0/1 and participates in the next comparison.

Final Answer:
1