What is the value of x after executing the following statement?
int x = 011 | 0x10;
Answer: C
Correct answer: 25
The two literals are in different bases: a leading 0 marks octal and 0x marks hexadecimal.
The bitwise OR sets every bit that is set in either, giving \(11001_2 = 25\).
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What is the output of this C code?
void main()
{
int a = 5, b = -7, c = 0, d;
d = ++a && ++b || ++c;
printf("\n%d%d%d%d", a, b, c, d);
}
Answer: D
Correct answer: 6 -6 0 1
Evaluate left to right, remembering that any non-zero value is true.
++a makes a 6, which is true.++b makes b -6 - non-zero, so also true.&& is therefore true, so || short-circuits and ++c never runs: c stays 0.d gets the value of the whole expression, 1.Output: 6 -6 0 1.
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Which is not a storage class?
Answer: B
Correct answer: Struct
The storage classes in C are auto, register, static, extern and typedef - typedef being grouped with them by the grammar even though it names a type rather than allocating storage. struct is not among them: it introduces a type, and a structure variable still needs a storage class of its own.
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For the following statement find the values generated for p and q?
int p = 0, q = 1;
p = q++;
p = ++q;
p = q--;
p = --q;
Value of p & q are
Answer: A
Correct answer: 1 1
Follow the four statements from p = 0, q = 1:
p = q++; - p gets the old 1; q becomes 2.p = ++q; - q becomes 3 and p gets 3.p = q--; - p gets the old 3; q becomes 2.p = --q; - q becomes 1 and p gets 1.So both end at 1.
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What is the output of this C code?
void main()
{
int a = -5;
int k = (a++, ++a);
printf("%d\n", k);
}
Answer: A
Correct answer: -3
The comma operator evaluates its left operand, discards the result, then evaluates the right operand and yields that. So a++ takes a to -4, then ++a takes it to -3 and produces -3. k is therefore -3. There is a sequence point at the comma, so this is well defined.
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What is the output of this C code?
int main()
{
int x = -2;
x = x >> 1;
printf("%d\n", x);
}
Answer: B
Correct answer: -1
-2 is all ones followed by a zero: \(\ldots 1110\). Shifting right on a signed negative value is an arithmetic shift in every mainstream implementation, so the sign bit is copied in from the left and the result is \(\ldots 1111 = -1\). Arithmetically it is a division by two, rounding toward negative infinity.
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What is the output of this C code?
void main()
{
int x = 4.3 % 2;
printf("Value of x is %d", x);
}
Answer: D
Correct answer: Compile time error
The % operator is defined only for integer operands. 4.3 is a double, so 4.3 % 2 is rejected at compile time - there is no implicit conversion to rescue it. For a floating-point remainder you would call fmod(4.3, 2) from <math.h>.
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What is the type of the below assignment expression if x is of type float, y is of type int?
y = x + y;
Answer: A
Correct answer: int
The type of an assignment expression is the type of its left operand, after the right-hand side has been converted to it. Here x + y is computed as a float, then converted to int for the assignment to y - so the expression y = x + y has type int, and the fractional part is lost.
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What will be the output of the following program?
#include <stdio.h>
Void f(static int*, extern int);
Static int b = 12;
Int main()
{
Static int a[5];
Register int I;
For(I = 0; I < 2> A[i++] = 2 * i++;
F(a, b);
For(I = 0; I < 2> Printf(“%d”, b++);
Return 0;
}
Void f (static int *x, extern int y)
{
Register int I;
For(I = 0; I < 2> *(++x +1) + = 2;
Y + = 2;
}
Answer: C
Correct answer: 0 12, though the listing this depends on has been destroyed
The program text in this record cannot be trusted. Every identifier has been capitalised - Void, Static, Int, For, Printf - which alone would stop it compiling, since C is case sensitive; the quotation marks in the printf call have become typographic quotes; and each of the three for loops has lost its body, ending at I < 2> where the loop condition, the increment and the statement should be. That last damage is characteristic: a stray < in stored HTML swallows everything up to the next >, and here it has taken the working part of every loop with it. One thing can still be said about the original. It declares void f(static int*, extern int);, and neither of those storage classes may be applied to a function parameter - C permits only register there - so the program the question was built on would not have compiled either. The keyed answer is “0 12”, which is consistent with a program whose array elements are never usefully assigned and whose file-scope b is unchanged by the call, since y += 2 inside f alters only the local copy of the argument. Beyond that the item cannot be worked until the original listing is restored.
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What is the value of X in the sample code given below?
double X; X = ( 2 + 3) * 2 + 3;
Answer: B
Correct answer: 13
The parentheses force the addition first, then multiplication precedes the final addition:
\[ (2 + 3) \times 2 + 3 = 5 \times 2 + 3 = 10 + 3 = 13 \]X is a double, so it stores 13.0, but the value is 13.
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