Expressions Questions and Answers

Q: 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); }

Correct answer: 6 -6 0 1Evaluate 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.The && 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.

Q: What is the correct and fully portable way to obtain the most significant byte of an unsigned integer x?

Correct answer: x > > ( CHAR_BIT * (sizeof(int) - 1))The most significant byte must be shifted down by the width of all the bytes below it. The portable way to say that is CHAR_BIT * (sizeof(int) - 1) - the number of bits in a byte times the number of bytes below the top one.x >> 24 assumes a 4-byte int and an 8-bit char. x & 0xFF00 masks the second byte, not the top one, and does not shift it down. The variant with sizeof(int) - 3 is simply the wrong count.

Q: In case of a conflict between the names of a local and global variable what happens?

Correct answer: The local variable is given a priority.The inner declaration shadows the outer one: within the block where the local is declared, the name refers to the local, and the global is simply not visible. It is not an error, and it does not depend on declaration order. The global can still be reached in C++ with ::name, but plain C offers no such escape.

Q: What is the output of this C code? int main() { int x = 2, y = 2; x /= x / y; printf("%d\n", x); return 0; }

Correct answer: 2x /= x / y; is x = x / (x / y). The right-hand side is computed first, with integer division throughout:\[ x / y = 2 / 2 = 1, \quad x = 2 / 1 = 2 \]So x is unchanged at 2.

Q: 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); }

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.

Q: The precedence of arithmetic operators is (from highest to lowest)?

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.

Q: What is the output of this C code? void main() { int x = 97; int y = sizeof(x++); printf("X is %d", x); }

Correct answer: x is 97sizeof 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.)

Q: What is the output of this C code? int main() { if (~0 == 1) printf("yes\n"); else printf("no\n"); }

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.

Q: What is the output of this C code? int main() { int i = 0; int j = i++ + i; printf("%d\n", j); }

Correct answer: 1With 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.

101.

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);
    }

A.6 -6 0 0
B.6 -5 0 1
C.-6 -6 0 1
D.6 -6 0 1

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.
The && 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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102.

What is the correct and fully portable way to obtain the most significant byte of an unsigned integer x?

A.x & 0xFF00
B.x > > 24
C.x > > ( CHAR_BIT * (sizeof(int) - 3))
D.x > > ( CHAR_BIT * (sizeof(int) - 1))

Answer: D

Correct answer: x > > ( CHAR_BIT * (sizeof(int) - 1))

The most significant byte must be shifted down by the width of all the bytes below it. The portable way to say that is CHAR_BIT * (sizeof(int) - 1) - the number of bits in a byte times the number of bytes below the top one.

x >> 24 assumes a 4-byte int and an 8-bit char. x & 0xFF00 masks the second byte, not the top one, and does not shift it down. The variant with sizeof(int) - 3 is simply the wrong count.

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103.

In case of a conflict between the names of a local and global variable what happens?

A.The global variable is given a priority.
B.The local variable is given a priority.
C.Which one will get a priority depends upon which one is defined first.
D.The compiler reports an error.

Answer: B

Correct answer: The local variable is given a priority.

The inner declaration shadows the outer one: within the block where the local is declared, the name refers to the local, and the global is simply not visible. It is not an error, and it does not depend on declaration order. The global can still be reached in C++ with ::name, but plain C offers no such escape.

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104.

What is the output of this C code?

 int main()
    {
        int x = 2, y = 2;
        x /= x / y;
        printf("%d\n", x);
        return 0;
    }

A.2
B.1
C.0.5
D.Undefined behaviour

Answer: A

Correct answer: 2

x /= x / y; is x = x / (x / y). The right-hand side is computed first, with integer division throughout:

\[ x / y = 2 / 2 = 1, \quad x = 2 / 1 = 2 \]

So x is unchanged at 2.

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105.

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);
    }

A.6
B.5
C.Varies

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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106.

The precedence of arithmetic operators is (from highest to lowest)?

A.%, *, /, +, -
B.%, +, /, *, -
C.+, -, %, *, /
D.%, +, -, *, /

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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107.

What is the output of this C code?

void main()
    {
        int x = 97;
        int y = sizeof(x++);
        printf("X is %d", x);
    }

A.X is 97
B.X is 98
C.X is 99
D.Run time error

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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108.

What is the output of this C code?

   int main()
    {
        if (~0 == 1)
            printf("yes\n");
        else
            printf("no\n");
    }

A.yes
B.no
C.Compile time error
D.Undefined

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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109.

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;
}

A.The program will result in compile time error
B.The program will print 12.35 22.35
C.The program will print 13.35 22.35
D.The program will print 1.35 11.35

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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110.

What is the output of this C code?

int main()
    {
        int i = 0;
        int j = i++ + i;
        printf("%d\n", j);
    }

A.1
B.2
C.Compile time error

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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