dave@safari.UUCP (dave munroe) (01/28/88)
>someone awhile back said that the default cc compiles to the 68000 instruction >set and that you need to specify -68010 to get that instruction set. >However, i compiled a program both ways and the a.out was the same. >So whats the story do i need to use -68010 ????? The differences between the 68010 and 68000 are mainly in the areas of systems programming: - support for virtual memory (by stacking the machine state during a page fault and then continuing the instruction after the page has been brought in) - support for a virtual machine concept by making MOVE to/from SR privileged, plus MOVEC, MOVES, and MOVE from CCR. - additional registers for supervisor mode programming: the Vector Base Register and alternate function code registers (SFC, DFC). - different exception handling (e.g. for bus [addressing] errors) - a two-word (4-byte) prefetch queue and a special "loop mode" of operation. Loop mode is transparent to the programmer and is entered when certain common instructions are used with the DBcc instruction, e.g. here loop: move.w (a0)+,(a1)+ dbeq d0,loop the 68010 enters a loop mode. Looping in this mode is efficient since only operand fetches are performed until the exit condition is met (opcode fetches are eliminated since the move.w is held in the instruction decode register and the dbeq is held in the prefetch queue). So, for most programming situations, you are not likely to see any differences by specifying cc -68010. It would be nice if some compilers were smart enough to optimize code so that the loop mode could be taken advantage of, but I don't know of any that do. -dave
alex@umbc3.UMD.EDU (Alex S. Crain) (02/01/88)
In article <231@safari.UUCP> dave@safari.UUCP (dave munroe) writes: >>someone awhile back said that the default cc compiles to the 68000 >>instruction >>set and that you need to specify -68010 to get that instruction set. >>However, i compiled a program both ways and the a.out was the same. >>So whats the story do i need to use -68010 ????? > >The differences between the 68010 and 68000 are mainly in the areas of >systems programming: [diff -c 68000 68010 > /dev/null] A quick link of /bin/echo to /lib/ccom and /lib/optim (the peephole optimizer) shows that if you specify -68000 or -68010, NOTHING HAPPENS! the cc program simply eats the option and goes along its way. Dave is correct in saying that the usable differences in the instruction sets are small, basicly the dbxx and rtd instructions. The former is a tight loop mode that requires significant reworking if the code structure that must (must ?) be done BEFORE code generation, and therefore cannot be done in a peephole optimizer. The second is an extended rts instruction that allows faster returns from functions with a fixed # of arguments. Dave is also correct is sating that /lib/ccom has never heard of these instructions. SO, the answer to the first question (how to get 68010 code) is "you can't" (partially true) or "get another compiler". There are compilers that know about tight loops and rtd, such as gcc, the Free Software Foundation compiler, which does a great deal of front end optimization (fix the code before we emit it). Unfortunatly, These compilers are either expensive, or in the case of gcc, free but still under development. -- :alex. nerwin!alex@umbc3.umd.edu alex@umbc3.umd.edu
pjc@pcbox.UUCP (Paul J. Condie) (02/01/88)
In article <231@safari.UUCP> dave@safari.UUCP (dave munroe) writes: >>someone awhile back said that the default cc compiles to the 68000 instruction >>set and that you need to specify -68010 to get that instruction set. > >So, for most programming situations, you are not likely to see any differences >by specifying cc -68010. It would be nice if some compilers were smart enough >to optimize code so that the loop mode could be taken advantage of, but I >don't know of any that do. that's nice but my question was does cc default to 68000 or 68010?