MIPS Instruction Decoding
I'm trying to understand how to decode MIPS binary instructions.
I compiled a hello world program in C on a Debian MIPS system with gcc and objdump shows me that the first instruction in the .text section is:
600: 03e00025 move zero,ra
I don't understand how it determines that this is the MOVE instruction.
03e00025 is 00000011111000000000000000100101 in binary. If I understand correctly the first 6 bits here are the opcode, which in this case is all 0, meaning it's an R-type instruction, so we have to look at the last 6 bits, which is 100101. Looking at the MIPS Instruction Set Manual it looks like this should be the OR instruction. I can't even find MOVE in that manual.
Googling for this I found that apparently there are "pseudo" instructions in the assembly, and supposedly move $t, $s expands to addiu $t, $s, 0, but if I look in the manual ADDIU has opcode 001001. Another result I found claims it translates to ADD but the last six bits of ADD should be 100000, so that doesn't fit either.
What am I missing?
assembly mips elf disassembly objdump
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
asked Nov 23 '18 at 23:33
StefanStefan
60028
add a comment |
I'm trying to understand how to decode MIPS binary instructions.
I compiled a hello world program in C on a Debian MIPS system with gcc and objdump shows me that the first instruction in the .text section is:
600: 03e00025 move zero,ra
I don't understand how it determines that this is the MOVE instruction.
03e00025 is 00000011111000000000000000100101 in binary. If I understand correctly the first 6 bits here are the opcode, which in this case is all 0, meaning it's an R-type instruction, so we have to look at the last 6 bits, which is 100101. Looking at the MIPS Instruction Set Manual it looks like this should be the OR instruction. I can't even find MOVE in that manual.
Googling for this I found that apparently there are "pseudo" instructions in the assembly, and supposedly move $t, $s expands to addiu $t, $s, 0, but if I look in the manual ADDIU has opcode 001001. Another result I found claims it translates to ADD but the last six bits of ADD should be 100000, so that doesn't fit either.
What am I missing?
assembly mips elf disassembly objdump
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
asked Nov 23 '18 at 23:33
StefanStefan
60028
1
Yes that is encoded asor $0, $ra, $0.
– Jester
Nov 23 '18 at 23:58
add a comment |
I'm trying to understand how to decode MIPS binary instructions.
I compiled a hello world program in C on a Debian MIPS system with gcc and objdump shows me that the first instruction in the .text section is:
600: 03e00025 move zero,ra
I don't understand how it determines that this is the MOVE instruction.
03e00025 is 00000011111000000000000000100101 in binary. If I understand correctly the first 6 bits here are the opcode, which in this case is all 0, meaning it's an R-type instruction, so we have to look at the last 6 bits, which is 100101. Looking at the MIPS Instruction Set Manual it looks like this should be the OR instruction. I can't even find MOVE in that manual.
Googling for this I found that apparently there are "pseudo" instructions in the assembly, and supposedly move $t, $s expands to addiu $t, $s, 0, but if I look in the manual ADDIU has opcode 001001. Another result I found claims it translates to ADD but the last six bits of ADD should be 100000, so that doesn't fit either.
What am I missing?
assembly mips elf disassembly objdump
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
asked Nov 23 '18 at 23:33
StefanStefan
60028
I'm trying to understand how to decode MIPS binary instructions.
I compiled a hello world program in C on a Debian MIPS system with gcc and objdump shows me that the first instruction in the .text section is:
600: 03e00025 move zero,ra
I don't understand how it determines that this is the MOVE instruction.
03e00025 is 00000011111000000000000000100101 in binary. If I understand correctly the first 6 bits here are the opcode, which in this case is all 0, meaning it's an R-type instruction, so we have to look at the last 6 bits, which is 100101. Looking at the MIPS Instruction Set Manual it looks like this should be the OR instruction. I can't even find MOVE in that manual.
Googling for this I found that apparently there are "pseudo" instructions in the assembly, and supposedly move $t, $s expands to addiu $t, $s, 0, but if I look in the manual ADDIU has opcode 001001. Another result I found claims it translates to ADD but the last six bits of ADD should be 100000, so that doesn't fit either.
What am I missing?
assembly mips elf disassembly objdump
assembly mips elf disassembly objdump
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
asked Nov 23 '18 at 23:33
StefanStefan
60028
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
asked Nov 23 '18 at 23:33
StefanStefan
60028
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
edited Nov 23 '18 at 23:59
Peter Cordes
128k18190326
128k18190326
asked Nov 23 '18 at 23:33
StefanStefan
60028
asked Nov 23 '18 at 23:33
StefanStefan
60028
asked Nov 23 '18 at 23:33
StefanStefan
60028
60028
1
Yes that is encoded asor $0, $ra, $0.
– Jester
Nov 23 '18 at 23:58
add a comment |
1
Yes that is encoded asor $0, $ra, $0.
– Jester
Nov 23 '18 at 23:58
1
1
Yes that is encoded as
or $0, $ra, $0.– Jester
Nov 23 '18 at 23:58
Yes that is encoded as
or $0, $ra, $0.– Jester
Nov 23 '18 at 23:58
add a comment |
1 Answer
1
active
oldest
votes
MIPS machine code doesn't have a specific opcode for move, but for the convenience of humans, many assemblers support pseudo-instructions like li, la, and move that assemble to one or more real machine instructions. addiu is a common one.
It would be totally correct for objdump to decode the instruction as or $0, $ra, $0 (according to Jester) to show you how it's actually encoded.
For some purposes it makes sense for a disassembler to decode any of the commonly used ways to copy a register into a move mnemonic. Adding or ORing an immediate 0, or a zero from reading the $zero register, do nothing to the value so it's copied unchanged.
When reading the asm, you normally don't care whether it's or, ori, addiu $0, $ra, 0, or whatever.
Different assemblers might use different implementations for the move pseudo-instruction, or hand-written asm could use any of them. I don't think there are any performance consequences either way. So the details of which machine instruction is used to implement a move depends on the assembler.
I'm not sure what the point of a move with a destination of $zero is. That would be a no-op, because $zero discards writes. (It's the CPU-register equivalent of /dev/zero)
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
add a comment |
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1 Answer
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votes
MIPS machine code doesn't have a specific opcode for move, but for the convenience of humans, many assemblers support pseudo-instructions like li, la, and move that assemble to one or more real machine instructions. addiu is a common one.
It would be totally correct for objdump to decode the instruction as or $0, $ra, $0 (according to Jester) to show you how it's actually encoded.
For some purposes it makes sense for a disassembler to decode any of the commonly used ways to copy a register into a move mnemonic. Adding or ORing an immediate 0, or a zero from reading the $zero register, do nothing to the value so it's copied unchanged.
When reading the asm, you normally don't care whether it's or, ori, addiu $0, $ra, 0, or whatever.
Different assemblers might use different implementations for the move pseudo-instruction, or hand-written asm could use any of them. I don't think there are any performance consequences either way. So the details of which machine instruction is used to implement a move depends on the assembler.
I'm not sure what the point of a move with a destination of $zero is. That would be a no-op, because $zero discards writes. (It's the CPU-register equivalent of /dev/zero)
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
add a comment |
MIPS machine code doesn't have a specific opcode for move, but for the convenience of humans, many assemblers support pseudo-instructions like li, la, and move that assemble to one or more real machine instructions. addiu is a common one.
It would be totally correct for objdump to decode the instruction as or $0, $ra, $0 (according to Jester) to show you how it's actually encoded.
For some purposes it makes sense for a disassembler to decode any of the commonly used ways to copy a register into a move mnemonic. Adding or ORing an immediate 0, or a zero from reading the $zero register, do nothing to the value so it's copied unchanged.
When reading the asm, you normally don't care whether it's or, ori, addiu $0, $ra, 0, or whatever.
Different assemblers might use different implementations for the move pseudo-instruction, or hand-written asm could use any of them. I don't think there are any performance consequences either way. So the details of which machine instruction is used to implement a move depends on the assembler.
I'm not sure what the point of a move with a destination of $zero is. That would be a no-op, because $zero discards writes. (It's the CPU-register equivalent of /dev/zero)
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
add a comment |
MIPS machine code doesn't have a specific opcode for move, but for the convenience of humans, many assemblers support pseudo-instructions like li, la, and move that assemble to one or more real machine instructions. addiu is a common one.
It would be totally correct for objdump to decode the instruction as or $0, $ra, $0 (according to Jester) to show you how it's actually encoded.
For some purposes it makes sense for a disassembler to decode any of the commonly used ways to copy a register into a move mnemonic. Adding or ORing an immediate 0, or a zero from reading the $zero register, do nothing to the value so it's copied unchanged.
When reading the asm, you normally don't care whether it's or, ori, addiu $0, $ra, 0, or whatever.
Different assemblers might use different implementations for the move pseudo-instruction, or hand-written asm could use any of them. I don't think there are any performance consequences either way. So the details of which machine instruction is used to implement a move depends on the assembler.
I'm not sure what the point of a move with a destination of $zero is. That would be a no-op, because $zero discards writes. (It's the CPU-register equivalent of /dev/zero)
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
MIPS machine code doesn't have a specific opcode for move, but for the convenience of humans, many assemblers support pseudo-instructions like li, la, and move that assemble to one or more real machine instructions. addiu is a common one.
It would be totally correct for objdump to decode the instruction as or $0, $ra, $0 (according to Jester) to show you how it's actually encoded.
For some purposes it makes sense for a disassembler to decode any of the commonly used ways to copy a register into a move mnemonic. Adding or ORing an immediate 0, or a zero from reading the $zero register, do nothing to the value so it's copied unchanged.
When reading the asm, you normally don't care whether it's or, ori, addiu $0, $ra, 0, or whatever.
Different assemblers might use different implementations for the move pseudo-instruction, or hand-written asm could use any of them. I don't think there are any performance consequences either way. So the details of which machine instruction is used to implement a move depends on the assembler.
I'm not sure what the point of a move with a destination of $zero is. That would be a no-op, because $zero discards writes. (It's the CPU-register equivalent of /dev/zero)
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
answered Nov 24 '18 at 0:08
Peter CordesPeter Cordes
128k18190326
128k18190326
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
add a comment |
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
Oh wow, awesome answer! Now I get it :-) Thank you very much!
– Stefan
Nov 24 '18 at 0:47
add a comment |
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Yes that is encoded as
or $0, $ra, $0.– Jester
Nov 23 '18 at 23:58