What is the output of this C code?
int main()
{
int a = 20, b = 15, c = 5;
int d;
d = a == (b + c);
printf("%d", d);
}
Answer: A
Correct answer: 1
= has lower precedence than ==, so the statement is d = (a == (b + c)). Since \(15 + 5 = 20\) and a is 20, the comparison is true, and a true relational expression in C has the value 1.
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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: A
Correct answer: 4
<< shifts the bits left, and each place shifted left doubles the value:
So x becomes 4. Shifting a small positive int left is well defined, so the result does not depend on the compiler or the machine.
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Which of the following is an invalid assignment operator?
Answer: D
Correct answer: None of the mentioned
All three are valid compound assignment operators: %= for remainder, /= for division and |= for bitwise OR. So none of them is invalid, and the answer is none of the mentioned.
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What will be the output of the following program?
#include <stdio.h>
static int y = 1;
int main()
{
static int z;
printf("%d %d", y, z);
return 0;
}
Answer: C
Correct answer: 1 0
y is a file-scope static int explicitly initialised to 1. z is a static local with no initialiser, and every object with static storage duration is zero-initialised - unlike an auto variable, which would hold junk. So the output is 1 0.
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What is the output of this C code?
int main()
{
int a = 10, b = 5, c = 3;
b != !a;
c = !!a;
printf("%d\t%d", b, c);
}
Answer: A
Correct answer: 5 1
b != !a; is an expression statement whose value is computed and thrown away - it does not assign anything, so b remains 5. c = !!a applies logical NOT twice: !10 is 0 and !0 is 1, so the double negation reduces any non-zero value to 1. That idiom is the usual way of normalising a value to 0 or 1.
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What is the output of this C code?
void main()
{
1 < 2 ? return 1: return 2;
}
Answer: D
Correct answer: Compile time error
return is a statement, not an expression, so it cannot appear as a branch of the conditional operator. 1 < 2 ? return 1: return 2; is therefore a syntax error, and the program never runs. The valid way to write it is return 1 < 2 ? 1 : 2;.
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What will be the output of the following program?
#include <stdio.h>
int main()
{
register int i = 2;
static char ch = 'A';
auto float j;
int k;
k = ++ch && i;
k = ++ch;
j = i-- + ++k * 2;
printf("%d %f", k, j);
return 0;
}
Answer: C
Correct answer: 68 138.000000
Follow the four statements, with ch starting at 'A' = 65.
k = ++ch && i; - ch becomes 66; both operands are non-zero, so k is 1.k = ++ch; - ch becomes 67 and k takes that value.j = i-- + ++k * 2; - i-- contributes 2 (leaving i at 1); ++k makes k 68, and \(68 \times 2 = 136\); so \(j = 2 + 136 = 138\).Printed with %d and %f: 68 138.000000.
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Which of the following statement is correct about the code snippet given below?
#include <stdio.h>
int main()
{
int a = 10, b = 2, c;
a = !( c = c == c) && ++b;
c += ( a + b--);
printf( " %d %d %d", b, c, a);
return 0;
}
Answer: A
Correct answer: The program will print the output 1 3 0
Take the statements in order.
a = !(c = c == c) && ++b; - c == c is 1, so c becomes 1 and !1 is 0. The && short-circuits, so ++b never runs and b stays 2. a becomes 0.c += (a + b--); - a + b is \(0 + 2 = 2\), so c becomes \(1 + 2 = 3\), and the post-decrement leaves b at 1.Printing b, c, a gives 1 3 0.
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What will be the output of the following code?
static int i = 5;
main()
{
int sum = 0;
do
{
sum += (1/i);
}while(0 < i--);
printf("sum of the series is %d", sum);
}
Answer: C
Correct answer: It will produce a run time error.
The loop is do { sum += 1/i; } while (0 < i--);. Because the test is 0 < i--, the last successful test happens when i is 1, and the post-decrement then leaves i at 0 - so the body runs once more and evaluates 1/0. Integer division by zero is a run-time fault, not something the compiler can catch, so the program crashes. (Had it survived, integer division would have made every term but 1/1 zero, so sum would have been 1, not the series in the first option.)
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What is the difference between the following 2 codes?
//Program 1
int main()
{
int d, a = 1, b = 2;
d = a++ + ++b;
printf("%d %d %d", d, a, b);
}
//Program 2
int main()
{
int d, a = 1, b = 2;
d = a++ + ++b;
printf("%d %d %d", d, a, b);
}
Answer: A
Correct answer: No difference as space doesn’t make any difference, values of a, b, d are same in both the case
The two programs are character for character the same, so naturally they produce the same values: a++ contributes 1 and leaves a as 2, ++b makes b 3 and contributes 3, so d is 4 and the output is 4 2 3. The point the question is driving at is that whitespace between tokens is insignificant - though it is worth knowing the limit of that rule: a+++++b would not compile, because the compiler takes the longest possible token at each step and reads it as a++ ++ +b.
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