Bitwise operation returning a long ... how

[DaTurk]

int number = 0;
int result = 0;

result = number | 0x80000000

Why is the compiler telling me this is converting a long to an int.
Where's the long? 0x80000000 is 32 bits, what am I missing?

[=?ISO-8859-1?Q?Arne_Vajh=F8j?=]

DaTurk wrote:

int number = 0;
int result = 0;

result = number | 0x80000000

Why is the compiler telling me this is converting a long to an int.
Where's the long? 0x80000000 is 32 bits, what am I missing?

0x80000000 is an uint (not an int).

And apperently int | uint gives a long.

Arne

[Michael C]

"Arne Vajhøj" <ar**@vajhoej.dkwrote in message
news:46***********************@news.sunsite.dk...

>Why is the compiler telling me this is converting a long to an int.
Where's the long? 0x80000000 is 32 bits, what am I missing?

0x80000000 is an uint (not an int).

And apperently int | uint gives a long.

This is a bit of a pita in C# imo when defining constants etc.

Michael

[Ben Voigt [C++ MVP]]

"DaTurk" <mm******@hotmail.comwrote in message
news:11**********************@57g2000hsv.googlegro ups.com...

int number = 0;
int result = 0;

result = number | 0x80000000

Why is the compiler telling me this is converting a long to an int.
Where's the long? 0x80000000 is 32 bits, what am I missing?

If you think that's fun, try:

byte a, b, c;
a = 1;
b = 5;
c = a & b;