Exercise: Expressions

Questions for: Expressions

What will be the output of the program? 
#include<stdio.h>
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i || ++j && ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}
A:
2, 2, 0, 1
B:
1, 2, 1, 0
C:
-2, 2, 0, 0
D:
-2, 2, 0, 1
Answer: D

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i || ++j && ++k; here (++j && ++k;) this code will not get executed because ++i has non-zero value.
becomes m = -2 || ++j && ++k;
becomes m = TRUE || ++j && ++k; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of variable 'i' only increemented by '1'(one). The variable j,k are not increemented.

Hence the output is "-2, 2, 0, 1".

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Assuming, integer is 2 byte, What will be the output of the program? 
#include<stdio.h>

int main()
{
    printf("%x\n", -2<<2);
    return 0;
}
A:
ffff
B:
0  
C:
fff8
D:
Error
Answer: C
The integer value 2 is represented as 00000000 00000010 in binary system.

Negative numbers are represented in 2's complement method.

1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).

2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).

Therefore, in binary we represent -2 as: 11111111 11111110.

After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.
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What will be the output of the program? 
#include<stdio.h>
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j && ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}
A:
-2, 3, 1, 1
B:
2, 3, 1, 2
C:
1, 2, 3, 1
D:
3, 3, 1, 2
Answer: A

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).

Hence the output is "-2, 3, 1, 1".

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In which order do the following gets evaluated
1. Relational
2. Arithmetic
3. Logical
4. Assignment
A:
2134
B:
1234
C:
4321
D:
3214
Answer: A

2. Arithmetic operators: *, /, %, +, -
1. Relational operators: >, <, >=, <=, ==, !=
3. Logical operators : !, &&, ||
4. Assignment operators: =

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Which of the following are unary operators in C?
1. !
2. sizeof
3. ~
4. &&
A:
1, 2
B:
1, 3
C:
2, 4
D:
1, 2, 3
Answer: D

An operation with only one operand is called unary operation.
Unary operators:
! Logical NOT operator.
~ bitwise NOT operator.
sizeof Size-of operator.

&& Logical AND is a logical operator.

Therefore, 1, 2, 3 are unary operators.

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