What is the value of x in this C code?
void main()
{
int x = 4 *5 / 2 + 9;
}
Answer: C
Correct answer: 19
* and / bind tighter than + and, having equal precedence, associate left to right:
Every step is integer arithmetic, so the answer is exactly 19.
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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 output of this C code?
void main()
{
int y = 3;
int x = 7 % 4 * 3 / 2;
printf("Value of x is %d", x);
}
Answer: A
Correct answer: Value of x is 4
%, * and / all have the same precedence and associate left to right, so the expression is grouped as ((7 % 4) * 3) / 2:
The final division is integer division, so the 0.5 is discarded and x is 4. The declaration of y is never used.
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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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The precedence of arithmetic operators is (from highest to lowest)?
Answer: A
Correct answer: %, *, /, +, -
%, * and / sit at the same, higher level and associate left to right; + and - sit at the same, lower level. The listed order reflects that grouping. So in a + b * c the multiplication happens first, while in a % b * c the two are simply taken left to right.
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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 data type will throw an error on modulus operation(%)?
Answer: C
Correct answer: float
% requires integer operands. char, short and int are all integer types and are promoted to int before the operation, so they are fine; float is not, and using it gives a compile-time error. Use fmod() for a floating-point remainder.
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What is the output of this C code?
int main()
{
int a = 20;
double b = 15.6;
int c;
c = a + b;
printf("%d", c);
}
Answer: A
Correct answer: 35
a + b promotes the int to double, giving \(20 + 15.6 = 35.6\). Assigning that to an int truncates toward zero rather than rounding, so c becomes 35. Rounding would have given 36 - the distractor.
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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?
void main()
{
int x = 0;
if (x = 0)
printf("Its zero\n");
else
printf("Its not zero\n");
}
Answer: A
Correct answer: Its not zero
if (x = 0) uses the assignment operator, not ==. It assigns 0 to x, and the value of the assignment expression is that same 0, which is false. So the else branch runs and it prints Its not zero - the classic typo that compilers now warn about.
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