Functions Questions and Answers

Answer: C

The keyed result is a compilation error.

A function declared void promises to return no value. The C standard makes a return statement with an expression in a void function a constraint violation, so the compiler must issue a diagnostic, and C++ compilers and strict C compilers reject it outright.

The value is not printed automatically, since returning a value never prints anything. The problem is caught when the program is compiled, not when it runs, so 'runtime error' is wrong; and the compiler does not silently ignore it.

Answer: A

ackermann(2,3) returns 16.

With i = 1 the function doubles repeatedly: ackermann(1, j) = \(2^{j}\), since ackermann(1,1) = 2 and each step applies \(2 \times\) to the previous value. Now unwind ackermann(2,3).

\[ \begin{aligned} a(2,1) &= 2 \\ a(2,2) &= a(1, a(2,1)) = a(1,2) = 4 \\ a(2,3) &= a(1, a(2,2)) = a(1,4) \\ &= 2^{4} = 16 \end{aligned} \]

Check a(1,4): it calls a(0, a(1,3)), and a(1,3) = 8, so the result is \(2 \times 8 = 16\). Options 20, 25 and 30 do not come from any correct reading of the recursion.

Answer: D

Computing fib(8) takes 41 calls in all.

Let C(n) be the number of invocations for fib(n), counting the call itself. fib(1) and fib(2) make one call each; any larger n makes one call plus the calls for n-1 and n-2.

\[ \begin{aligned} C(n) &= C(n-1) + C(n-2) + 1 \\ C(3) &= 3,\ C(4) = 5,\ C(5) = 9 \\ C(6) &= 15,\ C(7) = 25 \\ C(8) &= 25 + 15 + 1 = 41 \end{aligned} \]

A shortcut: \(C(n) = 2F(n) - 1\), and fib(8) = 21 gives \(2 \times 21 - 1 = 41\). 21 is the value fib(8) returns, not the number of calls, and 13 is fib(7).

Answer: A

The default parameter passing mechanism in languages such as C, C++ and Java is call by value: the value of the actual parameter is copied into the formal parameter, and changes made inside the function do not affect the caller. Call by reference, by address and by name need special syntax or are used in other languages, so they are not the default. The keyed answer is Call by Value.

Answer: C

The matching is a-ii, b-iii, c-iv, d-i. Call by reference is a way of passing arguments to a function, so it goes with function. The operator sizeof(float) gives 4, the number of bytes of a float. The symbol & as a binary operator is the bitwise AND. The statement return (a, b); is an incorrect statement in the sense that a function cannot return two values this way, as the comma operator would give only b. Option (A) pairs call by reference with an incorrect statement. Options (B) and (D) reverse several of the pairs. Only (C) fits all four.

Answer: A

Correct answer: Declared within the scope of a block, usually a function

An automatic variable is one declared inside a block. It comes into existence when the block is entered and disappears when it is left. Declaring it outside all functions would make it static; writing auto is optional, since it is the default.

Answer: D

Correct answer: Both (a) and (b)

An automatic variable is visible from its declaration to the end of the block that contains it, and that includes any blocks nested inside - unless one of them declares another variable of the same name, which shadows it.

Answer: A

Correct answer: True

C passes everything by value, but passing a pointer by value gives the function the caller’s address and so the effect of call by reference. Both styles are therefore available, which is why arrays - whose names decay to pointers - behave as though passed by reference.

Answer: D

Correct answer: 5

y holds the address of x, and p is set to that same address, so *p is x itself - 5. (The code stores an address in an unsigned int, which is poor practice and provokes warnings; it happens to work only where a pointer and an unsigned int are the same width.)

Answer: B

Correct answer: Compile time error

Two faults. Standard C does not allow a function to be defined inside another, and m is defined inside main here. And n, which calls m, is never called at all - so even if it compiled, nothing would be printed.