Unary + is the only dummy operator in C. So it has no effect on the
expression and now the while loop is, while(i--!=0) which is false
and so breaks out of while loop. The value –1 is printed due to the postdecrement
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.
The expression can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
The boolean expression needs to be evaluated only till the truth value of
the expression is not known. j is not equal to zero itself means that the
expression’s truth value is 1. Because it is followed by || and true ||
(anything) => true where (anything) will not be evaluated. So the
remaining expression is not evaluated and so the value of i remains the
Similarly when && operator is involved in an expression, when any of the
operands become false, the whole expression’s truth value becomes false
and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.