NoWhereMan
Registered
I was wondering if MacOS 10 still uses the same crappy memory block previous versions had?
For those who don't know,
Unlike PCs, Macintosh's do not leave anything in memory (which is actually pretty good). Unfortunately unless memory in the same physical space (side by side) is available then programs cannot load, despite the fact that the amount of total memory available could support the program.
Let me try to explain a bit better:
Let say you have 14MB of memory
You load a 6MB program, a 2 meg program, and a 4 meg program
You have 2MB left
And your memory looks like this
-6MB(program1)- -2MB(program2)- -4MB(program3)- -2MB(free)-
If you unload the 2MB (program2) and try to load a 4MB program, it will not load because the 2 memory blocks (of 2MB each), are not side by side. It will state the memory is not available.
However, if you exit your 4MB(program3) and then load it back up and try the 2nd 4MB program it will load at that point.
Does MacOS X still have this bug/feature?
For those who don't know,
Unlike PCs, Macintosh's do not leave anything in memory (which is actually pretty good). Unfortunately unless memory in the same physical space (side by side) is available then programs cannot load, despite the fact that the amount of total memory available could support the program.
Let me try to explain a bit better:
Let say you have 14MB of memory
You load a 6MB program, a 2 meg program, and a 4 meg program
You have 2MB left
And your memory looks like this
-6MB(program1)- -2MB(program2)- -4MB(program3)- -2MB(free)-
If you unload the 2MB (program2) and try to load a 4MB program, it will not load because the 2 memory blocks (of 2MB each), are not side by side. It will state the memory is not available.
However, if you exit your 4MB(program3) and then load it back up and try the 2nd 4MB program it will load at that point.
Does MacOS X still have this bug/feature?