why this would result in long integer overflow

Question

I checked the document that long= int64 has range more than 900,000,000,000,000

Here is my code:

int r = 99;
long test1 = r*r*r         


        

Answer 1

The problem is that the literal "99" is being treated as an int. If you add "L" it will treat it as a long. To fix your compilation problem:

long test2 = 99L * 99L * 99L * 99L * 99L;

And to fix the "incorrect result" caused by integer overflow:

long r = 99;
long test1 = r * r * r * r * r;

The key point is that the expression to the right of the "=" is evaluated before the assignment to long r is done.

There are other literal suffixes you might be interested in:

Type    Suffix    Example
uint    U or u    100U
long    L or l    100L
ulong   UL or ul  100UL
float   F or f    123.45F
decimal M or m    123.45M

@m.edmonson, regarding your question about why it comes out to 919965907. What's happening, is that the value is "wrapping" around int.MaxValue. You can see this with a little test program:

int i = 99; // 99
i *= 99;    // 9801
i *= 99;    // 970299
i *= 99;    // 96059601
i *= 99;    // 919965907        should be 9509900499 but comes out to 919965907
            //                      which is (9509900499 % int.MaxValue)

long k = 9509900499 % int.MaxValue;

What is meant by "wrapping around"? When you exceed int.MaxValue by 1, the value "goes back" to int.MinValue.

int j = int.MaxValue;
j++;

bool isNowMinValue = (j == int.MinValue);   // true, the value has "wrapped around"

This is a bit simplistic; if you search for "integer overflow" you will get a better explanation. It's worth understanding how integers (and other numeric types) are represented with 32 bits:

http://en.wikipedia.org/wiki/Signed_number_representations