C Constants Questions and Answers

Answer: C

Correct answer: exchange (swap) the contents of s and b

Trace the three assignments with initial values \( s \) and \( b \):

\[ b \leftarrow s+b, \qquad s \leftarrow (s+b)-s = b, \qquad b \leftarrow (s+b)-b = s \]

so the two values are exchanged without a temporary. The technique is a curiosity rather than good practice: the intermediate sum can overflow, and a compiler produces better code from an ordinary swap using a temporary variable.

Answer: D

Correct answer: Error, because a constant variable cannot be changed

The declaration int const k = 11; makes k a const-qualified object, and k++ attempts to modify it, which the compiler rejects.

Note that int const and const int are equivalent - the qualifier may be written on either side of the type - so there is nothing wrong with the combination itself. It is the increment that fails. The error is a compile-time diagnostic, not a run-time fault, since const-ness is enforced entirely by the compiler.

Answer: B

Correct answer: a followed by a newline; b is never read

The trap is that the newline is itself a character in the input stream. Typing a, Enter, b, Enter puts four characters into stdin: ’a’, ’\n’, ’b’, ’\n’. getchar reads one character at a time and does not skip whitespace, so the first putchar(getchar()) reads and prints a, and the second reads the newline and prints it. The b is still sitting in the buffer when the program ends. The output is therefore the letter a followed by a line break - not ab, which would require the input to be typed as a single line ab, and not a b, which would need a space between them. This is the reason scanf("%d") followed by getchar() so often appears to skip an input: the newline left behind by the earlier read is consumed by the later one. The keyed option in this record read “this can’t be the input”, which describes nothing; its text has been corrected to the actual output.