((u32) (0-1) << (24)) ?

[ogedezvoltare]

I am study a source code, and i passed over this expression, I know what it does, but i want to know why the expression wasn't write in this form ((u32) 1 << (24)).
Thx

#typedef unsigned long u32

[JosAH]

I am study a source code, and i passed over this expression, I know what it does, but i want to know why the expression wasn't write in this form ((u32) 1 << (24)).
Thx

#typedef unsigned long u32

Those two expressions aren't the same: (assuming a 32 bit unsigned long)
the first one equals 0xff000000L while the second one is 0x01000000L.

The '-' (minus) operator has higher precedence than the << (shift left) operator.
0-1 causes wrap around to all bits 1. The expression could also be written as:

1) ~0<<24
2) 0xff<<24

and I'm sure many more ways exist to write this expression.

kind regards,

Jos

[ogedezvoltare]

Can you please explain to me the processing of the expression, like is done from the compiler, and can you use please numbers in base 2. I want to know all the operations that are done, and their precedence, before the final result comes out.
Thx