long long int ans = a*b VS long long int ans = (long long int) a*b

Question

I've written a code:

int a = 1000000000, b = 1000000000;
long long int ans = a * b;
cout << ans << '\n';

this code is causing overflow. I understand that a * b is causing the problem but I have taken long long int variable to keep a*b. But look at the following code:

int a = 1000000000, b = 1000000000;
long long int ans = (long long int)a * b;
cout << ans << '\n';

it's working fine causing no overflow. Does it make any temporary variable to hold the value when calculating? Please explain the reason behind this strange overflowing.

Answer 1

This makes two temporary variables, (long long int)a and (long long int)b. The second conversion is implicit.

Actual compilers might not bother, if the hardware has a 32*32->64 multiply, but officially the conversions have to occur. On 64 bits hardware, it's essentially free when you load an int in a 64 bit register.