What is the output of this C code?
int main()
{
int x = 2, y = 2;
x /= x / y;
printf("%d\n", x);
return 0;
}
Answer: A
Correct answer: 2
x /= x / y; is x = x / (x / y). The right-hand side is computed first, with integer division throughout:
So x is unchanged at 2.
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What is the output of this C code?
void main()
{
int x = 1, y = 0, z = 5;
int a = x && y || z++;
printf("%d", z);
}
Answer: A
Correct answer: 6
&& binds tighter than ||, so the expression is (x && y) || z++. With y zero the left side is false, so || must evaluate its right operand to decide the result - z++ runs and z becomes 6.
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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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What is the output of this C code?
void main()
{
int x = 97;
int y = sizeof(x++);
printf("X is %d", x);
}
Answer: A
Correct answer: x is 97
sizeof is evaluated at compile time from the type of its operand, and its operand is not executed. So x++ never runs, y becomes sizeof(int), and x is still 97. (Variable-length arrays are the one exception where the operand is evaluated.)
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What is the output of this C code?
int main()
{
if (~0 == 1)
printf("yes\n");
else
printf("no\n");
}
Answer: B
Correct answer: no
~0 flips every bit of zero, giving all ones - which in two’s complement is -1, not 1. So the comparison ~0 == 1 is false and the else branch prints no. Confusing ~ (bitwise NOT) with ! (logical NOT) is what the question is testing.
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Which of the following statement is correct about the code snippet given below?
#include <stdio.h>
int main()
{
float z = 12.35, c = 10;
if( ++z -z)
c += z;
else
c -= z;
printf( "%f %f", z, c);
return 0;
}
Answer: A
Correct answer: The program will result in compile time error
The statement is written c - = z; with a space between the two characters. C reads - and = as two separate tokens there, so the line parses as c - (= z), which is a syntax error. A compound assignment operator has to be written as one token: c -= z;.
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What is the output of this C code?
int main()
{
int i = 0;
int j = i++ + i;
printf("%d\n", j);
}
Answer: A
Correct answer: 1
With i starting at 0, i++ contributes 0 and leaves i as 1, and the second operand then reads 1, giving \(0 + 1 = 1\). Strictly this expression reads i in one operand while modifying it in the other with no sequence point between, which the standard leaves undefined; 1 is what mainstream compilers produce, but the expression should be split into two statements.
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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 = 1, b = 1, d = 1;
printf("%d, %d, %d", ++a + ++a+a++, a++ + ++b, ++d + d++ + a++);
}
Answer: A
Correct answer: 15, 4, 5
++a + ++a + a++ modifies a three times in one expression, and the three printf arguments may be evaluated in any order - the standard does not fix it. The expression modifies a variable more than once, and reads it as well, with no sequence point in between. The C standard makes that undefined behaviour, so the answer listed is what one common compiler happens to produce - another compiler, or another optimisation level, may give something else, and no program should depend on it. The lesson to take away is not the number but the rule: one variable, one modification per expression.
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What will be the output of the following code snippet?
Y = 5;
if (! Y > 10)
X = Y + 3;
else
X = Y + 10;
printf(" X = %d Y = %d", X, Y);
Answer: A
Correct answer: The program will print X = 15 Y = 5
! binds far tighter than >, so the condition is (!Y) > 10, not !(Y > 10). With Y equal to 5, !Y is 0, and \(0 > 10\) is false. The else branch runs, giving \(X = 5 + 10 = 15\), and Y is never modified, so the output is X = 15 Y = 5.
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