Necessity of J vs. JAL (and JR vs. JALR) in MIPS assembly
Formalizing the comments into an answer
J
/JR
can be emulated with JAL
/JALR
as the latter performs a super-set of the operations of the former.
As @Jester pointed out, routines (functions in C jargon) must be careful to preserve their return address present in $ra
.
Unless the routine is a leaf routine (one that doesn't do any call) $ra
must be saved somewhere anyway.
Actually both JAL
/JALR
and J
/JR
can be implemented one in terms of the other:
Emulate
JAL
/JALR
withJ
/JR
Original Emulated jal foo la $ra, ret_label j foo ret_label:
Emulate
J
/JR
withJAL
/JALR
Original Emulated j foo prolog: addi $sp, $sp, -4 sw $ra, ($sp) jal foo epilog: lw $ra, ($sp) addi $sp, $sp, 4
For this to work, the code must return to
epilog
. It is assumed that$ra
is mostly preserved in routines (hence the names of the labels). Many thanks to @EOF for point out a mistake in this snippet.
As @Peter pointed out, the access to the $pc
leads to an easier (for humans) emulation of JAL
/JALR
.
As @EOF pointed out, some RISC machine actually have only one instruction for JAL
/JALR
and J
/JR
given their innate entanglement.
Considering that jumps and call happen very often in a typical program, being able to implement easily (and executing them fast) is mandatory for any successful ISA.
Colin Weltin-Wu
Updated on August 22, 2022Comments
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Colin Weltin-Wu over 1 year
I signed up because I've been googling forever for an answer to this question and can't find one.
I'd like to know if the jump instructions WITHOUT linking are strictly necessary in MIPS?
I can imagine for example that using "AL" versions when not required would incur some power penalty, but is there any situation (that's not completely contrived or could be coded around relatively simply) where ONLY J/JR would work?
Thank you!
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EOF about 7 yearsThe emulation of
J
/JR
doesn't work that way, since the control flow doesn't return, so the stack adjustment is not undone by this code. Nevertheless, +1. -
Margaret Bloom about 7 years@EOF Yes, of course! Brainfart. I meant something like 1) save
$ra
2) do the jumps withjal
/jalr
3) restore$ra
. Fixing the answer. Thanks -
EOF about 7 yearsYeah, basically just treat this function like a non-leaf function regardless of whether it actually is a leaf function.
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Colin Weltin-Wu about 7 yearsThanks this makes perfect sense. To take this even further, I'd be curious to the impact of removing one pair versus the other in terms of code size/speed.
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Colin Weltin-Wu about 7 yearsFor example, if I were to remove J/JR, the compiler might be able to rearrange the code to use branches. I realize this is very pedantic, just curiosity.
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EOF about 7 years@ColinWeltin-Wu: There's another aspect to having a clear distinction between function calls and jumps: A function call gives extra information about future jumps to the processor (namely, the
return
from the called function). Without this information,return
is a fairly expensive operation, since it's an absolute branch, which means branch target prediction gets involved. OTOH, if the processor keeps around a small hardware stack of return addresses, the branch target is known well in advance, allowing prefetching and pipelining.