Given i = 0, j = 1, k = –1 x = 0.5, y = 0.0 What is the output of the…

Concept:

In C, every relational operator (<, >, ==, !=, <=, >=) evaluates to the int value 1 when the comparison is true and 0 when it is false. The logical OR operator (||) also produces only 0 or 1, and it short-circuits: once its left operand is already nonzero (true), the right operand is never evaluated and the result is 1 regardless of that operand's own value. Operator precedence also governs how a mixed expression is grouped: the arithmetic operators (*, +) bind before the relational operator (<), which in turn binds before the logical operator (||).

Application:

1. By operator precedence, the expression x * y < i + j || k groups as ((x * y) < (i + j)) || k.

2. Compute the arithmetic sub-expressions: x * y = 0.5 * 0.0 = 0.0, and i + j = 0 + 1 = 1.

3. Evaluate the relational comparison: 0.0 < 1 is true, which C represents as the integer 1.

4. Evaluate the logical OR: because the left operand (1) is already nonzero/true, || short-circuits — it does not evaluate k — so the whole expression evaluates to 1.

Cross-check:

Even if k were evaluated, its value (-1) is also nonzero/true, so 1 || (-1) would still be 1 — the short-circuit does not change the outcome here. This also confirms that || in C can only ever produce 0 or 1, never any other magnitude such as 2 or a negative number.

Answer: 1