→In the option B, -10%-2 gives remainder of “0” and then expression becomes 4/0/3.
→4/0 gives divide by zero error.

Bitwise AND is a binary operator that operates on two equal-length bit patterns. If both bits in the compared position of the bit patterns are 1, the bit in the resulting bit pattern is 1, otherwise 0.
Bitwise OR is also a binary operator that operates on two equal-length bit patterns, similar to bitwise AND. If both bits in the compared position of the bit patterns are 0, the bit in the resulting bit pattern is 0, otherwise 1.
Bitwise XOR also takes two equal-length bit patterns. If both bits in the compared position of the bit patterns are 0 or 1, the bit in the resulting bit pattern is 0, otherwise 1.

In general, a-=b means a=a-b. So from this we can conclude that correct option is A.

‘*’ and ‘%’ operator have the same precedence but the associativity is from left to write. So x = 3*4%5 = 12%5 = 2.

int i=5;
printf("%d%d%d", i++, i ,++i);
In printf() the firstly the calculation is done from right to left and then printing is done from left to right.
So first ++i will increase the value of i by 1 and make it 6. Now the i will be read simply for second calculation.And now i++ will be calculated which will increment the value of i by 1 and make it 7, but since it is post increment so first 6 will be printed and then increment will be done and then for rest two 7 and 7 will be printed .Hence the output will be 6,7,7.

According to given conditions, if x value is < = 25 means it will return true and for other values it will return false.
Alone y<=100 won't return true, it should be along with x<=25

Except !(num-65) can be replaced by num!=65

Step-1: Evaluate x ​ * 3 because multiplication has more priority than remaining operators x * 3→ 1.5
Step-2: && is logical AND. Both the statements are TRUE, then it returns 1 otherwise 0.
1.5 && 3 is TRUE. So, it returns 1.
Step-3: j | k is bitwise OR operator. It returns -1.
Step-4: ((x * 3) && 3) || (j | k) islogical OR operator. It returns 1
Note: The precedence is ((x * 3) && 3) || (j | k)

Step-1: Here, we are using post increment for both x and y. In post increment the value assign first and update after assigning.
x= 35 + 20
y= 37 + 56 (In this statement, ‘y’ will become 36 before performing pre increment)
Step-2: Final x=56 and y=93
Note: They given wrong options. We are added correct option.
Excluded for evaluation.

The above program we are using post increment operators. In post increment the value assign first and update after assigning.
main( )
{
char *str = “abcde”;
printf (“%c”, *str); /* It returns character ‘a’ */
printf (“%c”, *str++); /* It returns character ‘a’ because it is used post increment*/
printf (“%c”, *(str++)); /* It returns character ‘b’ because the previous increment is update here and present also they given post increment, so it only returns ‘b’ */
printf (“%s”, str); /* It returns aabcde */
}