Ll denom = 1ll; what does this mean?

#1

this 1ll. how does this work?

#2

hello @JaveedYara

by default any number is treated as int datatype type.
so here 1 is also acting as int.
adding that ll at the end will typecast it to long long .

but here that ll at the end will not create any difference.

ll denom=1 ;
is same as ll denom=1ll;

both are same.

to understand the difference.

print these two statements,

cout<< (1<<40); // overflow will occur

cout<<(( 1 ll)<<40 ) ; // overflow will not occur