Expressions Questions and Answers

Answer: A

Correct answer: x is 97

sizeof is evaluated at compile time from the type of its operand, and its operand is not executed. So x++ never runs, y becomes sizeof(int), and x is still 97. (Variable-length arrays are the one exception where the operand is evaluated.)

Answer: B

Correct answer: no

~0 flips every bit of zero, giving all ones - which in two’s complement is -1, not 1. So the comparison ~0 == 1 is false and the else branch prints no. Confusing ~ (bitwise NOT) with ! (logical NOT) is what the question is testing.

Answer: A

Correct answer: The program will result in compile time error

The statement is written c - = z; with a space between the two characters. C reads - and = as two separate tokens there, so the line parses as c - (= z), which is a syntax error. A compound assignment operator has to be written as one token: c -= z;.

Answer: A

Correct answer: 1

With i starting at 0, i++ contributes 0 and leaves i as 1, and the second operand then reads 1, giving \(0 + 1 = 1\). Strictly this expression reads i in one operand while modifying it in the other with no sequence point between, which the standard leaves undefined; 1 is what mainstream compilers produce, but the expression should be split into two statements.

Answer: A

Correct answer: 35

a + b promotes the int to double, giving \(20 + 15.6 = 35.6\). Assigning that to an int truncates toward zero rather than rounding, so c becomes 35. Rounding would have given 36 - the distractor.

Answer: A

Correct answer: 15, 4, 5

++a + ++a + a++ modifies a three times in one expression, and the three printf arguments may be evaluated in any order - the standard does not fix it. 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. The lesson to take away is not the number but the rule: one variable, one modification per expression.

Answer: A

Correct answer: The program will print X = 15 Y = 5

! binds far tighter than >, so the condition is (!Y) > 10, not !(Y > 10). With Y equal to 5, !Y is 0, and \(0 > 10\) is false. The else branch runs, giving \(X = 5 + 10 = 15\), and Y is never modified, so the output is X = 15 Y = 5.

Answer: C

Correct answer: The code will result in L – value required

++num++ groups as ++(num++), because postfix binds tighter than prefix. The result of num++ is a value, not an object, so the prefix ++ has nothing it can modify - the compiler reports that an lvalue is required. The same applies to ++(a + b) or 5++.

Answer: C

Correct answer: -11

~ flips every bit, and for a two’s-complement integer that is exactly \(\sim a = -a - 1\):

The variable is unsigned, so it actually holds 4294967285, but %d reads those bits back as a signed int and prints -11.

Answer: B

Correct answer: For loop counter

register is worth asking for where a variable is accessed very often, and a loop counter is the classic case - it is read and written on every iteration. It cannot be used where the variable’s address is needed, since a register has no address. In practice modern compilers allocate registers better than the hint does, so the keyword is now of historical interest.