After executing the following instruction:
long time = 60 * 1000000000; // 60 seconds in nano-second units
What will be the value of the variable time ?
Your doing:
long x = (long)(int*int);
Do:
long x = long * long;
To prevent overflow of your ints
I was expecting a newbie to fall for it. ;D
Anyway you are absolutly right. This is what i used:
long time = 60L * 1000000000L;
But just as a curiosity C++ also messes up with our intuition on this one.
Oh crap, I better go change my code then 
well the 2nd number defaults to Long doesn’t it? :-*
1000000000 = fits perfectly in 32 bits
4294967296 = max for unsigned 32 bits
2147483647 = max for signed 32 bits
And long values never default to long.
6000000000 = invalid, compile-time error
6000000000L = valid
ah yes, horray for compile time checking.