Arithmetic operation resulted in an overflow. (Add

问题:

I can't understand this error:

In this call to method SetVolume, Volume = 2055786000 and size = 93552000. Volume is an Integer property, and size is also Integer, as you can see.

The class is a partial class of a dbml entity class, however this Volume property is NOT a column in the database, it exist only in the partial class, as a "business object property".

View Detail shows:

Data > Item : In order to evaluate an indexed property, the property must be qualified and the arguments must be explicitly supplied by the user.

What may cause this...?

回答1:

The maximum value of an integer (which is signed) is 2147483647. If that value overflows, an exception is thrown to prevent unexpected behavior of your program.

If that exception wouldn't be thrown, you'd have a value of -2145629296 for your Volume, which is most probably not wanted.

Solution: Use an Int64 for your volume. With a max value of 9223372036854775807, you're probably more on the safe side.

回答2:

int.MaxValue = 2147483647
2055786000 + 93552000 = 2149338000 > int.MaxValue

So you cannot store this number into an integer. You could use Int64 type which has a maximum value of 9,223,372,036,854,775,807.

回答3:

For simplicity I will use bytes:

byte a=250;
byte b=8;
byte c=a+b;

if a, b, and c were 'int', you would expect 258, but in the case of 'byte', the expected result would be 2 (258 & 0xFF), but in a Windows application you get an exception, in a console one you may not (I don't, but this may depend on IDE, I use SharpDevelop).

Sometimes, however, that behaviour is desired (e.g. you only care about the lower 8 bits of the result).

You could do the following:

byte a=250;
byte b=8;

byte c=(byte)((int)a + (int)b);

This way both 'a' and 'b' are converted to 'int', added, then casted back to 'byte'.

To be on the safe side, you may also want to try:

...
byte c=(byte)(((int)a + (int)b) & 0xFF);

Or if you really want that behaviour, the much simpler way of doing the above is:

unchecked
{
    byte a=250;
    byte b=8;
    byte c=a+b;
}

Or declare your variables first, then do the math in the 'unchecked' section.

Alternately, if you want to force the checking of overflow, use 'checked' instead.

Hope this clears things up.

Nurchi

P.S.

Trust me, that exception is your friend :)

回答4:

The result integer value is out of the range which an integer data type can hold.

Try using Int64

回答5:

Maximum value fo int is 2147483647, so 2055786000+93552000 > 2147483647 and it caused overflow

回答6:

2055786000 + 93552000 = 2149338000, which is greater than 2^31. So if you're using signed integers coded on 4 bytes, the result of the operation doesn't fit and you get an overflow exception.

回答7:

The maximum size for an int is 2147483647. You could use an Int64/Long which is far larger.

回答8:

This error occurred for me when a value was returned as -1.#IND due to a division by zero. More info on IEEE floating-point exceptions in C++ here on SO and by John Cook

For the one who has downvoted this answer (and did not specify why), the reason why this answer can be significant to some is that a division by zero will lead to an infinitely large number and thus a value that doesn't fit in an Int32 (or even Int64). So the error you receive will be the same (Arithmetic operation resulted in an overflow) but the reason is slightly different.