Mixing
Trouble understanding LEA assembly
0
I am very new to assembly, so I just want to make sure I am understanding whats happening in this code:
400610: 83 ff 1d cmp $0x1d,%edi
400613: 7f 0c jg 400621 <f1+0x11>
400615: 89 f8 mov %edi,%eax
400617: c1 e0 04 shl $0x4,%eax
40061a: 8d 04 f8 lea (%rax,%rdi,8),%eax
40061d: 8d 04 78 lea (%rax,%rdi,2),%eax
400620: c3 retq
400621: c1 ff 02 sar $0x2,%edi
400624: 8d 47 11 lea 0x11(%rdi),%eax
400627: c3 retq
From what I can see, there is a jump to 400621 but I am not sure what f1+0x11 signifies.
If it does not jump, it continues and shifts %eax to the left 4 (multiplies by 16), then performs eax = rax + rdi * 8, then eax = rax + rdi * 2? I am not sure what the purpose of doing that twice is.
If it does jump, it shifts %eax to the right 2 (divides by 4) and then I am not sure what (lea 0x11(%rdi),%eax) does.
Help would be appreciated, thank you!
assembly x86-64
edited Nov 21 '18 at 0:06
Jester
46.6k34682
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
|
show 2 more comments
0
I am very new to assembly, so I just want to make sure I am understanding whats happening in this code:
400610: 83 ff 1d cmp $0x1d,%edi
400613: 7f 0c jg 400621 <f1+0x11>
400615: 89 f8 mov %edi,%eax
400617: c1 e0 04 shl $0x4,%eax
40061a: 8d 04 f8 lea (%rax,%rdi,8),%eax
40061d: 8d 04 78 lea (%rax,%rdi,2),%eax
400620: c3 retq
400621: c1 ff 02 sar $0x2,%edi
400624: 8d 47 11 lea 0x11(%rdi),%eax
400627: c3 retq
From what I can see, there is a jump to 400621 but I am not sure what f1+0x11 signifies.
If it does not jump, it continues and shifts %eax to the left 4 (multiplies by 16), then performs eax = rax + rdi * 8, then eax = rax + rdi * 2? I am not sure what the purpose of doing that twice is.
If it does jump, it shifts %eax to the right 2 (divides by 4) and then I am not sure what (lea 0x11(%rdi),%eax) does.
Help would be appreciated, thank you!
assembly x86-64
edited Nov 21 '18 at 0:06
Jester
46.6k34682
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
1
The reason for twoleainstructions is that the effective addressing is limited so you can't achieve that with a single one. If you understand those ones, what problem do you have withlea 0x11(%rdi),%eax? That just doeseax=edi+0x11.
– Jester
Nov 21 '18 at 0:06
Ah, yes, the maximum number it can be is 8, I know that. So for lea, when a number is outside of the parentheses, it just adds it? If I hadlea 0x11(%rdi,%rsi,4),%eaxit would doeax = rsi * 4 + rdi + 0x11? Also forjg 400621 <f1+0x11>, 400621 signifies the address, but what does the second part signify?
– Andrew Zaw
Nov 21 '18 at 0:08
1
1) Yes. Not specific toleaof course, that's just the general form of an address. 2) The same thing. It's just a friendly service of your disassembler showing it as an offset from the previous symbol.f1is presumably the name of this function sof1+0x11=0x400610+0x11=0x400621
– Jester
Nov 21 '18 at 0:21
Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 '18 at 0:51
"the maximum number it can be is 8" - the address scale in 64b mode can be only 1, 2, 4 or 8. ... it's not about "min/max" value, but only these powers of two are available, "5" is invalid too ((,%edi,5)= error ). (but you can still useleato do multiplication by 5 likelea (%eax, %eax, 4), %eax=>eax = eax + eax*4 = eax*5)
– Ped7g
Nov 21 '18 at 13:08
|
show 2 more comments
0
0
0
I am very new to assembly, so I just want to make sure I am understanding whats happening in this code:
400610: 83 ff 1d cmp $0x1d,%edi
400613: 7f 0c jg 400621 <f1+0x11>
400615: 89 f8 mov %edi,%eax
400617: c1 e0 04 shl $0x4,%eax
40061a: 8d 04 f8 lea (%rax,%rdi,8),%eax
40061d: 8d 04 78 lea (%rax,%rdi,2),%eax
400620: c3 retq
400621: c1 ff 02 sar $0x2,%edi
400624: 8d 47 11 lea 0x11(%rdi),%eax
400627: c3 retq
From what I can see, there is a jump to 400621 but I am not sure what f1+0x11 signifies.
If it does not jump, it continues and shifts %eax to the left 4 (multiplies by 16), then performs eax = rax + rdi * 8, then eax = rax + rdi * 2? I am not sure what the purpose of doing that twice is.
If it does jump, it shifts %eax to the right 2 (divides by 4) and then I am not sure what (lea 0x11(%rdi),%eax) does.
Help would be appreciated, thank you!
assembly x86-64
edited Nov 21 '18 at 0:06
Jester
46.6k34682
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
I am very new to assembly, so I just want to make sure I am understanding whats happening in this code:
400610: 83 ff 1d cmp $0x1d,%edi
400613: 7f 0c jg 400621 <f1+0x11>
400615: 89 f8 mov %edi,%eax
400617: c1 e0 04 shl $0x4,%eax
40061a: 8d 04 f8 lea (%rax,%rdi,8),%eax
40061d: 8d 04 78 lea (%rax,%rdi,2),%eax
400620: c3 retq
400621: c1 ff 02 sar $0x2,%edi
400624: 8d 47 11 lea 0x11(%rdi),%eax
400627: c3 retq
From what I can see, there is a jump to 400621 but I am not sure what f1+0x11 signifies.
If it does not jump, it continues and shifts %eax to the left 4 (multiplies by 16), then performs eax = rax + rdi * 8, then eax = rax + rdi * 2? I am not sure what the purpose of doing that twice is.
If it does jump, it shifts %eax to the right 2 (divides by 4) and then I am not sure what (lea 0x11(%rdi),%eax) does.
Help would be appreciated, thank you!
assembly x86-64
assembly x86-64
edited Nov 21 '18 at 0:06
Jester
46.6k34682
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
edited Nov 21 '18 at 0:06
Jester
46.6k34682
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
edited Nov 21 '18 at 0:06
Jester
46.6k34682
edited Nov 21 '18 at 0:06
Jester
46.6k34682
edited Nov 21 '18 at 0:06
Jester
46.6k34682
46.6k34682
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
asked Nov 21 '18 at 0:03
Andrew ZawAndrew Zaw
605
605
1
The reason for twoleainstructions is that the effective addressing is limited so you can't achieve that with a single one. If you understand those ones, what problem do you have withlea 0x11(%rdi),%eax? That just doeseax=edi+0x11.
– Jester
Nov 21 '18 at 0:06
Ah, yes, the maximum number it can be is 8, I know that. So for lea, when a number is outside of the parentheses, it just adds it? If I hadlea 0x11(%rdi,%rsi,4),%eaxit would doeax = rsi * 4 + rdi + 0x11? Also forjg 400621 <f1+0x11>, 400621 signifies the address, but what does the second part signify?
– Andrew Zaw
Nov 21 '18 at 0:08
1
1) Yes. Not specific toleaof course, that's just the general form of an address. 2) The same thing. It's just a friendly service of your disassembler showing it as an offset from the previous symbol.f1is presumably the name of this function sof1+0x11=0x400610+0x11=0x400621
– Jester
Nov 21 '18 at 0:21
Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 '18 at 0:51
"the maximum number it can be is 8" - the address scale in 64b mode can be only 1, 2, 4 or 8. ... it's not about "min/max" value, but only these powers of two are available, "5" is invalid too ((,%edi,5)= error ). (but you can still useleato do multiplication by 5 likelea (%eax, %eax, 4), %eax=>eax = eax + eax*4 = eax*5)
– Ped7g
Nov 21 '18 at 13:08
|
show 2 more comments
1
The reason for twoleainstructions is that the effective addressing is limited so you can't achieve that with a single one. If you understand those ones, what problem do you have withlea 0x11(%rdi),%eax? That just doeseax=edi+0x11.
– Jester
Nov 21 '18 at 0:06
Ah, yes, the maximum number it can be is 8, I know that. So for lea, when a number is outside of the parentheses, it just adds it? If I hadlea 0x11(%rdi,%rsi,4),%eaxit would doeax = rsi * 4 + rdi + 0x11? Also forjg 400621 <f1+0x11>, 400621 signifies the address, but what does the second part signify?
– Andrew Zaw
Nov 21 '18 at 0:08
1
1) Yes. Not specific toleaof course, that's just the general form of an address. 2) The same thing. It's just a friendly service of your disassembler showing it as an offset from the previous symbol.f1is presumably the name of this function sof1+0x11=0x400610+0x11=0x400621
– Jester
Nov 21 '18 at 0:21
Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 '18 at 0:51
"the maximum number it can be is 8" - the address scale in 64b mode can be only 1, 2, 4 or 8. ... it's not about "min/max" value, but only these powers of two are available, "5" is invalid too ((,%edi,5)= error ). (but you can still useleato do multiplication by 5 likelea (%eax, %eax, 4), %eax=>eax = eax + eax*4 = eax*5)
– Ped7g
Nov 21 '18 at 13:08
1
1
The reason for two
lea instructions is that the effective addressing is limited so you can't achieve that with a single one. If you understand those ones, what problem do you have with lea 0x11(%rdi),%eax? That just does eax=edi+0x11.– Jester
Nov 21 '18 at 0:06
The reason for two
lea instructions is that the effective addressing is limited so you can't achieve that with a single one. If you understand those ones, what problem do you have with lea 0x11(%rdi),%eax? That just does eax=edi+0x11.– Jester
Nov 21 '18 at 0:06
Ah, yes, the maximum number it can be is 8, I know that. So for lea, when a number is outside of the parentheses, it just adds it? If I had
lea 0x11(%rdi,%rsi,4),%eax it would do eax = rsi * 4 + rdi + 0x11? Also for jg 400621 <f1+0x11>, 400621 signifies the address, but what does the second part signify?– Andrew Zaw
Nov 21 '18 at 0:08
Ah, yes, the maximum number it can be is 8, I know that. So for lea, when a number is outside of the parentheses, it just adds it? If I had
lea 0x11(%rdi,%rsi,4),%eax it would do eax = rsi * 4 + rdi + 0x11? Also for jg 400621 <f1+0x11>, 400621 signifies the address, but what does the second part signify?– Andrew Zaw
Nov 21 '18 at 0:08
1
1
1) Yes. Not specific to
lea of course, that's just the general form of an address. 2) The same thing. It's just a friendly service of your disassembler showing it as an offset from the previous symbol. f1 is presumably the name of this function so f1+0x11=0x400610+0x11=0x400621– Jester
Nov 21 '18 at 0:21
1) Yes. Not specific to
lea of course, that's just the general form of an address. 2) The same thing. It's just a friendly service of your disassembler showing it as an offset from the previous symbol. f1 is presumably the name of this function so f1+0x11=0x400610+0x11=0x400621– Jester
Nov 21 '18 at 0:21
Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 '18 at 0:51
Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 '18 at 0:51
"the maximum number it can be is 8" - the address scale in 64b mode can be only 1, 2, 4 or 8. ... it's not about "min/max" value, but only these powers of two are available, "5" is invalid too (
(,%edi,5) = error ). (but you can still use lea to do multiplication by 5 like lea (%eax, %eax, 4), %eax => eax = eax + eax*4 = eax*5)– Ped7g
Nov 21 '18 at 13:08
"the maximum number it can be is 8" - the address scale in 64b mode can be only 1, 2, 4 or 8. ... it's not about "min/max" value, but only these powers of two are available, "5" is invalid too (
(,%edi,5) = error ). (but you can still use lea to do multiplication by 5 like lea (%eax, %eax, 4), %eax => eax = eax + eax*4 = eax*5)– Ped7g
Nov 21 '18 at 13:08
|
show 2 more comments
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1
The reason for two
leainstructions is that the effective addressing is limited so you can't achieve that with a single one. If you understand those ones, what problem do you have withlea 0x11(%rdi),%eax? That just doeseax=edi+0x11.– Jester
Nov 21 '18 at 0:06
Ah, yes, the maximum number it can be is 8, I know that. So for lea, when a number is outside of the parentheses, it just adds it? If I had
lea 0x11(%rdi,%rsi,4),%eaxit would doeax = rsi * 4 + rdi + 0x11? Also forjg 400621 <f1+0x11>, 400621 signifies the address, but what does the second part signify?– Andrew Zaw
Nov 21 '18 at 0:08
1
1) Yes. Not specific to
leaof course, that's just the general form of an address. 2) The same thing. It's just a friendly service of your disassembler showing it as an offset from the previous symbol.f1is presumably the name of this function sof1+0x11=0x400610+0x11=0x400621– Jester
Nov 21 '18 at 0:21
Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 '18 at 0:51
"the maximum number it can be is 8" - the address scale in 64b mode can be only 1, 2, 4 or 8. ... it's not about "min/max" value, but only these powers of two are available, "5" is invalid too (
(,%edi,5)= error ). (but you can still useleato do multiplication by 5 likelea (%eax, %eax, 4), %eax=>eax = eax + eax*4 = eax*5)– Ped7g
Nov 21 '18 at 13:08