Trouble understanding LEA assembly

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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!

edited Nov 21 at 0:06

Jester

45.9k34381

asked Nov 21 at 0:03

Andrew Zaw

605

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 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 at 0:08

1

1) Yes. Not specific tolea 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 at 0:21

Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 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
2 days ago

|
show 2 more comments

up vote
0
down vote

favorite

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!

edited Nov 21 at 0:06

Jester

45.9k34381

asked Nov 21 at 0:03

Andrew Zaw

605

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 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 at 0:08

1

1) Yes. Not specific tolea 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 at 0:21

Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 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
2 days ago

|
show 2 more comments

up vote
0
down vote

favorite

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!

edited Nov 21 at 0:06

Jester

45.9k34381

asked Nov 21 at 0:03

Andrew 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

edited Nov 21 at 0:06

Jester

45.9k34381

asked Nov 21 at 0:03

Andrew Zaw

605

edited Nov 21 at 0:06

Jester

45.9k34381

asked Nov 21 at 0:03

Andrew Zaw

605

edited Nov 21 at 0:06

Jester

45.9k34381

edited Nov 21 at 0:06

Jester

45.9k34381

edited Nov 21 at 0:06

Jester

45.9k34381

asked Nov 21 at 0:03

Andrew Zaw

605

asked Nov 21 at 0:03

Andrew Zaw

605

asked Nov 21 at 0:03

Andrew Zaw

605

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 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 at 0:08

1

1) Yes. Not specific tolea 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 at 0:21

Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 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
2 days ago

|
show 2 more comments

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 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 at 0:08

1

1) Yes. Not specific tolea 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 at 0:21

Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 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
2 days ago

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 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 at 0:08

1) Yes. Not specific tolea 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 at 0:21

Ah, that makes sense, thank you very much!
– Andrew Zaw
Nov 21 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
2 days ago

|
show 2 more comments

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