What is the value of the following expression?
i = 1;
i << 1 < 1>
Answer: A
Correct answer: 2
<< shifts the bits left, and one place left doubles the value:
The value of the expression is therefore 2. Shifting cannot produce a negative result here, since the sign bit of a small positive int is nowhere near the top.
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Which of the following is not an arithmetic operation?
Answer: D
Correct answer: a != 50;
!= is a relational operator: it compares and yields 1 or 0, and does not modify a at all. The other three are compound assignment operators - a *= 20 means a = a * 20, and likewise for /= and %=.
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Which of the following statement is correct about the code snippet given below?
#include <stdio.h>
int main()
{
int n = 12, k;
printf("%d", (k = sizeof( n + 12.0))++);
return 0;
}
Answer: C
Correct answer: The code will result in a compile time error
(k = sizeof(n + 12.0))++ tries to increment the result of an assignment expression. That result is a value, not an object, so it is not a modifiable lvalue and the compiler rejects it. (Had the increment been dropped, sizeof(n + 12.0) would be 8 on a typical machine, since the int is promoted to double.)
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What is the output of this C code?
int main()
{
int i = 10;
int *p = &i;
printf("%d\n", *p++);
}
Answer: A
Correct answer: 10
++ binds tighter than *, but it is a post-increment, so *p++ means “dereference the current p, then advance p”. The value printed is therefore i itself, 10; the pointer moves on afterwards, to an address that is no longer valid.
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What is the output of this C code?
void main()
{
int a = 5;
int b = ++a + a++ + --a;
printf("Value of b is %d", b);
}
Answer: D
Correct answer: Undefined behaviour
b = ++a + a++ + --a; modifies a three times within one expression, with no sequence point between the changes. The standard leaves the result undefined - not merely unspecified - so any answer a particular compiler gives is accidental. This is why such expressions must be split into separate statements.
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What is the output of the following program?
#include <stdio.h>
int main()
{
static int a = 3;
printf("%d", a--);
return 0;
}
Answer: C
Correct answer: 3
a-- is a post-decrement: the current value is what the expression yields, and the decrement happens afterwards. So printf receives 3; a becomes 2 immediately after, but nothing prints it. The static qualifier matters only if the function is called again.
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Longevity of a variable refers to
Answer: A
Correct answer: The duration for which the variable retains a given value during the execution of a program.
Longevity (or lifetime, or storage duration) is about time - how long the variable’s storage exists and keeps its value. That is a different question from scope, which is about place: the part of the program in which the name is visible. A static local has a short scope but the longest possible lifetime.
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Which of the following statement are correct?
(i) The maximum value a variable can hold depends upon its storage class.
(ii) By default all variables enjoy a static storage class.
Answer: D
Correct answer: Both I & II are incorrect
(i) is false: the range of a variable depends on its type - char, int, long - not on where it is stored. A static int and an auto int hold exactly the same range. (ii) is false: the default for a variable declared inside a function is auto, not static; only file-scope variables have static storage duration.
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What is the final value of j in the below code?
int main()
{
int i = 0, j = 0;
if (i && (j = i + 10))
//do something
;
}
Answer: D
Correct answer: 0
&& short-circuits: since i is 0 the whole condition is already false, so the right operand (j = i + 10) is never evaluated and the assignment never happens. j keeps its initial value, 0. This guarantee is what makes idioms such as if (p && p->next) safe.
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What is the output of this C code?
void main()
{
int k = 8;
int x = 0 == 1 && k++;
printf("%d%d\n", x, k);
}
Answer: B
Correct answer: 0 8
&& short-circuits: if the left operand is false the right is never evaluated. 0 == 1 is false, so k++ never runs and k stays 8, while x gets the value 0. The output is therefore 0 8 - printed as 08, since the format string has no separator.
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