Expressions Questions and Answers

Answer: D

Correct answer: 0

&& short-circuits: since i is 0 the whole condition is already false, so the right operand (j = i + 10) is never evaluated and the assignment never happens. j keeps its initial value, 0. This guarantee is what makes idioms such as if (p && p->next) safe.

Answer: B

Correct answer: No

i++ yields the old value 1, which is true, so the right operand is evaluated - but by then the side effect has happened and i is 2. i == 1 is therefore false, the whole condition is false, and the program prints No. The && is a sequence point, so the increment is guaranteed complete before the comparison.

Answer: A

Correct answer: True

&& and || are sequence points: the left operand is fully evaluated, side effects and all, before the right operand is touched. That is exactly what makes short-circuiting meaningful and what lets if (n != 0 && total / n > 5) be safe.

Answer: A

Correct answer: True

Both && and || short-circuit, and the standard requires it rather than leaving it to the compiler. If the left operand of && is false, or of || is true, the result is already determined and the right operand is not evaluated at all - side effects in it simply do not happen.

Answer: B

Correct answer: 0 or 1

A relational or logical expression in C yields an int that is exactly 1 for true and 0 for false. That is stricter than the rule for testing a value, where any non-zero counts as true. So (3 > 2) is 1, never 3 and never some other positive number.

Answer: B

Correct answer: 1

== binds tighter than =, so the statement is d = ((b + c) == a). Since \(5 + 5 = 10\) equals a, the comparison is true and yields 1. Had = bound tighter the code would not even compile, since d = b + c is not something you can compare in that position.

Answer: A

Correct answer: 5 1

b != !a; is an expression statement whose value is computed and thrown away - it does not assign anything, so b remains 5. c = !!a applies logical NOT twice: !10 is 0 and !0 is 1, so the double negation reduces any non-zero value to 1. That idiom is the usual way of normalising a value to 0 or 1.

Answer: D

Correct answer: =

= is the assignment operator: it stores a value rather than comparing or combining truth values. != and == are relational, and || is logical. Mistaking = for == in a condition is the classic C bug.

Answer: C

Correct answer: TRUE 1 TRUE 2

Both comparisons read a in one operand while modifying it in the other, with no sequence point between, so both are formally undefined behaviour. The expression modifies a variable more than once, and reads it as well, with no sequence point in between. The C standard makes that undefined behaviour, so the answer listed is what one common compiler happens to produce - another compiler, or another optimisation level, may give something else, and no program should depend on it. What the question is really illustrating is that a == a-- and a == --a cannot be reasoned about at all - not that one is true and the other false.

Answer: A

Correct answer: structure

Relational operators work on scalars - arithmetic types and pointers - because comparison needs a single value to order. A struct is an aggregate of several members, so s1 < s2 has no meaning and is rejected; even == is not defined for structures, since padding bytes would make a byte-wise comparison unreliable. Comparing them means comparing their members one by one.