Expressions Questions and Answers

Answer: C

Correct answer: 1

Pointer subtraction returns the number of elements between the two pointers, not the number of bytes. k holds the value p had before the increment, and p++ moved p on by one int, so p - k is 1. The compiler divides the 4-byte gap by sizeof(int) for you.

Answer: B

Correct answer: -1

In C the result of % takes the sign of the dividend, and the language requires that (a/b)*b + a%b == a. Since -5 / 4 truncates toward zero to \(-1\), we need \(-1 \times 4 + r = -5\), giving \(r = -1\). So the program prints -1.

Answer: A

Correct answer: 2

The value of an assignment expression is the value assigned. foo() returns 2, so x becomes 2 and the expression (x = foo()) itself has the value 2. That value is then what the surrounding != 1 compares against - and the comparison, being a separate expression, would yield 1.

Answer: B

Correct answer: 6 5

num is 0, so num <= 0 is true and !(num <= 0) is 0. The || must therefore evaluate its right operand: ++z takes z to 4 and is non-zero, so the if branch runs.

In printf("%d %d ", ++num + z++, ++z) the arguments are evaluated right to left on the usual compilers: ++z makes z 5 and yields 5; then ++num gives 1 and z++ yields 5, so the first argument is 6. Output: 6 5. Note that the order of argument evaluation is unspecified, so this particular pairing is compiler-dependent.

Answer: D

Correct answer: c = 4;

<<= shifts left and assigns. Shifting left by one position doubles the value, because each bit moves to the next higher power of two:

So c becomes 4.

Answer: A

Correct answer: a=23 c=R q = 345 p = 345

Each variable prints its own value, and the storage classes affect only lifetime and placement, not the values: a is 23, and c printed with %c shows the character R. The two long variables both hold 345 and print with %ld as 345. The only real trap is the case of the letters in the options - the format string writes a lower-case a.

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.

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: C

Correct answer: a = 5, b = 9

if (a = 5) assigns 5 to a, and the value of that assignment, 5, is non-zero and so true. The body runs, taking b from 10 to 9. The printf then shows a as 5 and, because b-- is a post-decrement, the value passed is the current 9 - the decrement to 8 happens after the value has been handed over, and nothing prints afterwards.

Answer: C

Correct answer: float

% requires integer operands. char, short and int are all integer types and are promoted to int before the operation, so they are fine; float is not, and using it gives a compile-time error. Use fmod() for a floating-point remainder.