CODEWITHSUNDEEP
c++integer-overflow
samnaction
samnaction

Reputation: 1254

Is overflow happening here?

I couldn't get why this program is giving this output. Is overflow happening here ?

int _tmain(int argc, _TCHAR* argv[])
{
     __int64 fileAgeInFileTime ;
     fileAgeInFileTime = 24  *60 * 60 ;

     cout << fileAgeInFileTime << endl;

     fileAgeInFileTime *= 10000000;

     __int64 fileAgeInFileTime2 = 24  *60 * 60  * 10000000 ;

     cout << fileAgeInFileTime << "  " <<fileAgeInFileTime2;

    return 0;
}

O/P :

86400

864000000000 711573504

I couldnt understand why fileAgeInFileTime and fileAgeInFileTime2 have different values ?

The real requirement is to get the 100 nano second resolution of file.

Upvotes: 3

Views: 95

Answers (2)

Luk&#225;š Bednař&#237;k
Luk&#225;š Bednař&#237;k

Reputation: 2638

Yes, there's an overflow. You multiply four int constants and assign them to __int64.

__int64 can be saved inside long long (long long should have at least 64 bits since C++11). You need to mark your constants as long long like this:

__int64 fileAgeInFileTime2 = 24LL * 60LL * 60LL * 10000000LL;

Or mark the first of those constants as LL, the others are implicitly converted:

__int64 fileAgeInFileTime2 = 24LL * 60 * 60 * 10000000;

Some kind of cast also works (other variables are also implicitly converted):

__int64 fileAgeInFileTime2 = static_cast<__int64>(24) * 60 * 60 * 10000000;

The last line should also explain why

__int64 fileAgeInFileTime2 = 10000000;
fileAgeInFileTime2 *= 24 * 60 * 60;

works. This multiplies four ints, converts the result to int64 and multiplies it by fileAgeInFileTime2. The total result is then assign to fileAgeInFileTime2.

Upvotes: 3

alexeykuzmin0
alexeykuzmin0

Reputation: 6440

Yes, you have overflow here. When you multiply int by int, the result is also int, so you have overflow while calculating 24 * 60 * 60 * 1000000. After this the incorrect result is stored in __int64 variable.

Here type conversions for arithmetic operators are explained

Upvotes: 2

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