what is this assembly doing exactly?
I have an assembly code that I am reading, and there is the following part in the code:
mov $0x28,%esi
mov %rax,%rdi
callq <strchr@plt>
mov %rax,%rbx
test %rax,%rax
jne somewhere
I am having a bit of difficulty in this code. The first move moves 0x28 to register %esi. The second moves %rax to %rdi, and then it does strchr(char *rdi, char *esi). So, basically it returns the pointer to the first occurence after ascii 0x28, which is '('. Then last test tells me that it is testing whether or not string at rdi has a '(' character. If does not have, then test %rax, %rax will set ZeroFlags, and jne will not execute. Is my understanding correct?
One more question I have is when I try to check what exactly is in %esi after the first mov, I use the command p (char *)$esi in gdb but I get an output <error: Cannot access memory at address 0x28>. Could anyone advise me how I can view contents of $esi?
c assembly gdb
asked Nov 21 '18 at 20:13
dipankardipankar
294
add a comment |
I have an assembly code that I am reading, and there is the following part in the code:
mov $0x28,%esi
mov %rax,%rdi
callq <strchr@plt>
mov %rax,%rbx
test %rax,%rax
jne somewhere
I am having a bit of difficulty in this code. The first move moves 0x28 to register %esi. The second moves %rax to %rdi, and then it does strchr(char *rdi, char *esi). So, basically it returns the pointer to the first occurence after ascii 0x28, which is '('. Then last test tells me that it is testing whether or not string at rdi has a '(' character. If does not have, then test %rax, %rax will set ZeroFlags, and jne will not execute. Is my understanding correct?
One more question I have is when I try to check what exactly is in %esi after the first mov, I use the command p (char *)$esi in gdb but I get an output <error: Cannot access memory at address 0x28>. Could anyone advise me how I can view contents of $esi?
c assembly gdb
asked Nov 21 '18 at 20:13
dipankardipankar
294
2
Yes you got it correct.esiholds the character itself not a pointer to it. You can dop (char)$esiorp/c $esi.
– Jester
Nov 21 '18 at 20:15
2
It would be more likestrchr(char *rdi, int esi). Also the factesiis not a pointer is implicit,esiis only 32-bit.
– Havenard
Nov 21 '18 at 20:31
add a comment |
I have an assembly code that I am reading, and there is the following part in the code:
mov $0x28,%esi
mov %rax,%rdi
callq <strchr@plt>
mov %rax,%rbx
test %rax,%rax
jne somewhere
I am having a bit of difficulty in this code. The first move moves 0x28 to register %esi. The second moves %rax to %rdi, and then it does strchr(char *rdi, char *esi). So, basically it returns the pointer to the first occurence after ascii 0x28, which is '('. Then last test tells me that it is testing whether or not string at rdi has a '(' character. If does not have, then test %rax, %rax will set ZeroFlags, and jne will not execute. Is my understanding correct?
One more question I have is when I try to check what exactly is in %esi after the first mov, I use the command p (char *)$esi in gdb but I get an output <error: Cannot access memory at address 0x28>. Could anyone advise me how I can view contents of $esi?
c assembly gdb
asked Nov 21 '18 at 20:13
dipankardipankar
294
I have an assembly code that I am reading, and there is the following part in the code:
mov $0x28,%esi
mov %rax,%rdi
callq <strchr@plt>
mov %rax,%rbx
test %rax,%rax
jne somewhere
I am having a bit of difficulty in this code. The first move moves 0x28 to register %esi. The second moves %rax to %rdi, and then it does strchr(char *rdi, char *esi). So, basically it returns the pointer to the first occurence after ascii 0x28, which is '('. Then last test tells me that it is testing whether or not string at rdi has a '(' character. If does not have, then test %rax, %rax will set ZeroFlags, and jne will not execute. Is my understanding correct?
One more question I have is when I try to check what exactly is in %esi after the first mov, I use the command p (char *)$esi in gdb but I get an output <error: Cannot access memory at address 0x28>. Could anyone advise me how I can view contents of $esi?
c assembly gdb
c assembly gdb
asked Nov 21 '18 at 20:13
dipankardipankar
294
asked Nov 21 '18 at 20:13
dipankardipankar
294
asked Nov 21 '18 at 20:13
dipankardipankar
294
asked Nov 21 '18 at 20:13
dipankardipankar
294
asked Nov 21 '18 at 20:13
dipankardipankar
294
294
2
Yes you got it correct.esiholds the character itself not a pointer to it. You can dop (char)$esiorp/c $esi.
– Jester
Nov 21 '18 at 20:15
2
It would be more likestrchr(char *rdi, int esi). Also the factesiis not a pointer is implicit,esiis only 32-bit.
– Havenard
Nov 21 '18 at 20:31
add a comment |
2
Yes you got it correct.esiholds the character itself not a pointer to it. You can dop (char)$esiorp/c $esi.
– Jester
Nov 21 '18 at 20:15
2
It would be more likestrchr(char *rdi, int esi). Also the factesiis not a pointer is implicit,esiis only 32-bit.
– Havenard
Nov 21 '18 at 20:31
2
2
Yes you got it correct.
esi holds the character itself not a pointer to it. You can do p (char)$esi or p/c $esi.– Jester
Nov 21 '18 at 20:15
Yes you got it correct.
esi holds the character itself not a pointer to it. You can do p (char)$esi or p/c $esi.– Jester
Nov 21 '18 at 20:15
2
2
It would be more like
strchr(char *rdi, int esi). Also the fact esi is not a pointer is implicit, esi is only 32-bit.– Havenard
Nov 21 '18 at 20:31
It would be more like
strchr(char *rdi, int esi). Also the fact esi is not a pointer is implicit, esi is only 32-bit.– Havenard
Nov 21 '18 at 20:31
add a comment |
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2
Yes you got it correct.
esiholds the character itself not a pointer to it. You can dop (char)$esiorp/c $esi.– Jester
Nov 21 '18 at 20:15
2
It would be more like
strchr(char *rdi, int esi). Also the factesiis not a pointer is implicit,esiis only 32-bit.– Havenard
Nov 21 '18 at 20:31