🔒 Locked

Why are these the same?

Started by Esap1 Sep 16, 2000 at 11:39 PM 7 replies 250+ views

Original Post

Sep 16, 2000 at 11:39 PM

Why are these statements the same, They shouldnt be: __int64 ints,ints1; ints = (80 << 56) | (70 << 48) | (60 << 40) | (50 << 32) | (40 << 24) | (30 << 16) | (20 << 8) | 10; ints1 = (40 << 24) | (30 << 16) | (20 << 8) | 10; if(ints==ints1){WHY ARE THEY THE SAME????} They are the same, why, why would __int64 only hold 32 bits?

Esap1

Author

Esap1

Sep 17, 2000 at 1:30 AM

anyone?

Esap1

Author

Esap1

Sep 17, 2000 at 3:25 AM

Could someone atleast tell me there not SUPPOSED to be the same?

Metron

Sep 17, 2000 at 3:55 AM

Hi,

this sounds to me as if the == does not a 64 bit compare but only 32 bit. If you look at your line of code, you will see that the lower 32 bit of both __int64 are equal. I couldn''t find the definition of the compare operator... perhaps there isn''t one for 64 bit ?

Kind regards,
Metron

Ysaneya

208

Ysaneya

Sep 17, 2000 at 8:47 AM

I simply guess your compiler takes your integer constants (80,60,40..) as ints (4 bytes) as any other compiler. Just typecast everything to __int64 and i believe it''ll work.

Y.

NBGH

Sep 17, 2000 at 10:27 AM

Win32 does not support true 64bit data types, they cheat and truncate them to 32bit without you knowing. Trust me, I''ve done a lot of programming with storage media and I know because its a pain in the...

Esap1

Author

Esap1

Sep 17, 2000 at 1:56 PM

NBGH: So your saying it will never work?

Esap1

Author

Esap1

Sep 17, 2000 at 11:18 PM

Is there any way to get it 2 work?

DLife

Sep 18, 2000 at 12:44 AM

You must cast all calculation operands to __int64.

DLife

Topic Locked

This topic has been locked by a moderator. New replies are not allowed.