In C, evaluate expressions in printf arguments left to right: what prints? #include int main() { int x = 55; printf("%d, %d, %d ", x <= 55, x = 40, x >= 10); return 0; }

Introduction / Context:This question assesses your grasp of relational expressions, assignment expressions as r-values, and the order in which function arguments are evaluated in C. It also clarifies that printf simply prints the values of its evaluated arguments.

Given Data / Assumptions:

• x starts at 55.
• Expressions are evaluated as separate arguments before the call prints them.
• The assignment expression x = 40 yields the value 40.

Concept / Approach:Each argument is an expression. The first compares the original x with 55; the second assigns a new value to x; the third uses the updated x in another comparison. Many modern compilers evaluate function arguments left to right, and in this typical exam context the intent is precisely that sequence.

Step-by-Step Solution:Argument 1: x <= 55 → 55 <= 55 → true → prints 1.Argument 2: x = 40 → assignment returns 40 → prints 40; x is now 40.Argument 3: x >= 10 → 40 >= 10 → true → prints 1.Output: 1, 40, 1.

Verification / Alternative check:Manually break it into temporaries, then printf those temporaries to confirm the same result.

Why Other Options Are Wrong:(b) and (c) wrongly print the old value of x in the middle. (d) treats the middle as boolean rather than the assignment result. (e) incorrectly sets the first relation.

Common Pitfalls:Assuming assignments in printf are not evaluated or thinking that printf prints variable names rather than values. Also, confusing the printed integer 1/0 with true/false semantics.

Final Answer:1, 40, 1