long long int initialization warnings

Question

2 Questions

First, while

long long int num = 1000000000000;

works fine

long long int num = 4014109449;

gives

warning: this decimal constant is unsigned only in ISO C90 [enabled by default]

What does it mean ?

Secondly

long long int num = 1000000*1000000;

gives an overflow warning while

long long int num = 1000000000000;

is ok,even though they are same.How do i get rid of it? Multiplication gives a garbage value

Answer 1

The problem is that the value 4014109449 is an unsigned long int in C90 but a long long int in C99 because it is too large for a 32-bit long int. While 1000000000000 is too large for any 32-bit type, so is automatically a long long int. The warning relates to the fact that the behaviour differs between C90 and C99.

The solution is to force type agreement between the literal and the variable type by using an appropriate type suffix. In this case:

long long num = 4014109449LL ;

or use a type cast:

long long num = (long long)4014109449 ;

Similarly the expression 1000000 * 1000000 is a multiply of two int types and has an int result, but causes an overflow - there is no automatic promotion to a larger type for int expressions. The solution is again to be explicit about the type of the literal:

long long num = 1000000LL * 1000000LL;

or you can also use a type cast on one or both operands.

long long num = (long long)1000000 * 1000000;