Expressions Questions and Answers

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.

Q: 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); }

Correct answer: 35a + 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.

Q: 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++); }

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.

Q: 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);

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.

Q: Which of the following statement is correct about the code snippet given below? num = 5; printf( "%d", ++num++ );

Correct answer: The code will result in L – value required++num++ groups as ++(num++), because postfix binds tighter than prefix. The result of num++ is a value, not an object, so the prefix ++ has nothing it can modify - the compiler reports that an lvalue is required. The same applies to ++(a + b) or 5++.

Q: What is the output of this C code? int main() { unsigned int a = 10; a = ~a; printf("%d\n", a); }

Correct answer: -11~ flips every bit, and for a two’s-complement integer that is exactly \(\sim a = -a - 1\):\[ \sim 10 = -10 - 1 = -11 \]The variable is unsigned, so it actually holds 4294967285, but %d reads those bits back as a signed int and prints -11.

Q: For which of the following situation should the register storage class be used?

Correct answer: For loop counterregister is worth asking for where a variable is accessed very often, and a loop counter is the classic case - it is read and written on every iteration. It cannot be used where the variable’s address is needed, since a register has no address. In practice modern compilers allocate registers better than the hint does, so the keyword is now of historical interest.

Q: p++ executes faster than p + 1 since

Correct answer: Single machine instruction required for p++p++ maps to a single increment instruction that updates p in place, whereas p + 1 computes a new value that then has to be stored somewhere. The saving is historical, though: any modern optimising compiler generates identical code for p = p + 1 and p++.

Q: Comment on the output of this C code? int main() { int i, n, a = 4; scanf("%d", &n); for (i = 0; i < n; i++) a = a * 2; }

Correct answer: No outputThe loop multiplies a by 2 on each of the n passes, so a ends as \(4 \times 2^n\). But the program never prints anything - there is no printf after the loop - so the computed value is simply discarded when main returns. No output.

Q: Where will the space be allocated for an automatic storage class variable?

Correct answer: In memoryAn automatic variable lives on the stack, which is memory. The register class is the one that asks for a CPU register, and even that is only a request. A compiler may of course keep an automatic variable in a register as an optimisation, but the storage class itself specifies memory.

111.

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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112.

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);
    }

A.35
B.36
C.35.6
D.30

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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113.

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++);
    }

A.15, 4, 5
B.9, 6, 9
C.9, 3, 5
D.6, 4, 6

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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114.

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);

A.The program will print X = 15 Y = 5
B.The program will print X = 15 Y = 0
C.The program will print X = 8 Y = 5
D.The program will print X = 3 Y = 0

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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115.

Which of the following statement is correct about the code snippet given below?

num = 5;
printf( "%d", ++num++ );

A.The code will print 5
B.The code will print 6
C.The code will result in L - value required
D.The code will result in R - value required

Answer: C

Correct answer: The code will result in L – value required

++num++ groups as ++(num++), because postfix binds tighter than prefix. The result of num++ is a value, not an object, so the prefix ++ has nothing it can modify - the compiler reports that an lvalue is required. The same applies to ++(a + b) or 5++.

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116.

What is the output of this C code?

int main()
    {
        unsigned int a = 10;
        a = ~a;
        printf("%d\n", a);
    }

A.-9
B.-10
C.-11
D.10

Answer: C

Correct answer: -11

~ flips every bit, and for a two’s-complement integer that is exactly \(\sim a = -a - 1\):

\[ \sim 10 = -10 - 1 = -11 \]

The variable is unsigned, so it actually holds 4294967285, but %d reads those bits back as a signed int and prints -11.

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117.

For which of the following situation should the register storage class be used?

A.For local variable in a function
B.For loop counter
C.For collecting values returned from a function
D.For variables used in a recursive function

Answer: B

Correct answer: For loop counter

register is worth asking for where a variable is accessed very often, and a loop counter is the classic case - it is read and written on every iteration. It cannot be used where the variable’s address is needed, since a register has no address. In practice modern compilers allocate registers better than the hint does, so the keyword is now of historical interest.

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118.

p++ executes faster than p + 1 since

A.P uses registers
B.Single machine instruction required for p++
C.Option a and b
D.None

Answer: B

Correct answer: Single machine instruction required for p++

p++ maps to a single increment instruction that updates p in place, whereas p + 1 computes a new value that then has to be stored somewhere. The saving is historical, though: any modern optimising compiler generates identical code for p = p + 1 and p++.

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119.

Comment on the output of this C code?

int main()
    {
        int i, n, a = 4;
        scanf("%d", &n);
        for (i = 0; i < n; i++)
            a = a * 2;
    }

A.Logical Shift left
B.No output
C.Arithmetic Shift right
D.bitwise exclusive OR

Answer: B

Correct answer: No output

The loop multiplies a by 2 on each of the n passes, so a ends as \(4 \times 2^n\). But the program never prints anything - there is no printf after the loop - so the computed value is simply discarded when main returns. No output.

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120.

Where will the space be allocated for an automatic storage class variable?

A.In CPU register
B.In memory as well as in CPU register
C.In memory
D.On disk

Answer: C

Correct answer: In memory

An automatic variable lives on the stack, which is memory. The register class is the one that asks for a CPU register, and even that is only a request. A compiler may of course keep an automatic variable in a register as an optimisation, but the storage class itself specifies memory.

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