Linked by Thom Holwerda on Sat 3rd Feb 2018 14:15 UTC, submitted by Drumhellar
Mac OS X

When users attempt to launch a 32-bit app in 10.13.4, it will still launch, but it will do so with a warning message notifying the user that the app will eventually not be compatible with the operating system unless it is updated. This follows the same approach that Apple took with iOS, which completed its sunset of 32-bit app support with iOS 11 last fall.

This is good. I would prefer other companies, too, take a more aggressive approach towards deprecating outdated technology in consumer technology.

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The Myth of 64 bits
by softdrat on Sun 4th Feb 2018 05:00 UTC
softdrat
Member since:
2008-09-17

64 is twice 32, so it must be better. Right?

NOT!

Most of the apps that I run are written by .. me. All working fine in a 32 bit environment.

So what happens when you move to a 64 bit machine? Gah! Memory exhaustion. All those 32 bit pointers, which were largely innocous before, now become 64 bit pointers, for which almost half of the bits are zero. So if our app uses lots of pointers, like mine do, now the memory footprint has nearly doubled, all in order to store zeroes.

At least fore me, going to 64 bits means that nearly half your memery is wasted. YMMV

Reply Score: 2

RE: The Myth of 64 bits
by ahferroin7 on Mon 5th Feb 2018 12:53 in reply to "The Myth of 64 bits"
ahferroin7 Member since:
2015-10-30

So you're seriously dealing with enough pointers that it actually matters that they double in length on a platform that pretty much universally has at least 4G of RAM these days?

If that's the case, then you probably need to be re-evaluating how your code is written, as it can almost certainly be made far more memory efficient. Even the Linux kernel doesn't have double the memory footprint when built 64-bit that it does when built 32-bit, and it uses pointers all over the place and it has a very large number of other data structures that are larger on a 64-bit kernel.

Reply Parent Score: 3

RE: The Myth of 64 bits
by grat on Mon 5th Feb 2018 19:43 in reply to "The Myth of 64 bits"
grat Member since:
2006-02-02

I presume this means you're storing a "char" (8 bit) as a 64 bit value?

If your code doesn't distinguish between bytes, short and long ints, perhaps you should put the keyboard away before you hurt someone.

Unless your code is spewing off pointers by the bucket, the extra 4 bytes per pointer shouldn't be costing you more than a kilobyte or two (256 pointers will cost you 1k. Are you really using thousands of variables in your code??).

Reply Parent Score: 2