Assigning value to array element in assembly (GAS syntax)
0
I have the following C code that returns 9:
#include <stdio.h>
int main()
{
int array[4]={0};
array[2]=9;
return array[2];
}
To accomplish the same in assembly language I have tried:
.section .data
array: .zero 4
.section .text
.globl _start
_start:
mov $2,%esi
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx
mov $9,%ecx
mov $1,%eax
mov array(,%esi,4),%ebx # return the value of array[2]
int $0x80
assembled and linked with:
gcc -g -c array.s && ld array.o && ./a.out
However this program returns 0 instead of 9:
>./a.out
>echo $?
0
I suspect the problem is in the lines with comments. After various unsuccessful attempts to find the problem I decided to ask the question: How to change of the value an array element (in this case array[2]) in assembly?
assembly gas att
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
add a comment |
0
I have the following C code that returns 9:
#include <stdio.h>
int main()
{
int array[4]={0};
array[2]=9;
return array[2];
}
To accomplish the same in assembly language I have tried:
.section .data
array: .zero 4
.section .text
.globl _start
_start:
mov $2,%esi
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx
mov $9,%ecx
mov $1,%eax
mov array(,%esi,4),%ebx # return the value of array[2]
int $0x80
assembled and linked with:
gcc -g -c array.s && ld array.o && ./a.out
However this program returns 0 instead of 9:
>./a.out
>echo $?
0
I suspect the problem is in the lines with comments. After various unsuccessful attempts to find the problem I decided to ask the question: How to change of the value an array element (in this case array[2]) in assembly?
assembly gas att
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
2
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx- this loads value from memory. To get address in similar way one can uselea array(,%esi,4),%ecx. Andmov $9,%ecxloads value 9 intoecx, it does not involve any extra memory access except the instruction opcode itself. And your finalmov array(,%esi,4),%ebxis completely correct and doing what it should, but if you will check how it differs from that "move address to ecx", there's no difference, so I'm not sure why you expect different result. (also using debugger to watch CPU state per instruction is of tremendous help)
– Ped7g
Nov 23 '18 at 15:50
1
And to modify array in fixed way, you can skip all those extra preparations, and just do immediatelymovl $9, array+2*4to use theMOV r/m32, imm32instruction to write immediate directly into memory. If you insist on indexing (to learn it for later), thenmovl $9, array(,%esi,4)writes value 9 into memory at "array + esi*4". To use "address" inecx(if it would hold one), you can domovl $9,(%ecx), to write into memory, the parentheses make it memory operand, not register, as target.
– Ped7g
Nov 23 '18 at 15:54
Thanks, now I see the flaws with the code I posted. Bothmovl $9, array+2*4andmovl $9,array(,%esi,4)give the desired result.
– Sergio
Nov 23 '18 at 16:46
add a comment |
0
0
0
I have the following C code that returns 9:
#include <stdio.h>
int main()
{
int array[4]={0};
array[2]=9;
return array[2];
}
To accomplish the same in assembly language I have tried:
.section .data
array: .zero 4
.section .text
.globl _start
_start:
mov $2,%esi
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx
mov $9,%ecx
mov $1,%eax
mov array(,%esi,4),%ebx # return the value of array[2]
int $0x80
assembled and linked with:
gcc -g -c array.s && ld array.o && ./a.out
However this program returns 0 instead of 9:
>./a.out
>echo $?
0
I suspect the problem is in the lines with comments. After various unsuccessful attempts to find the problem I decided to ask the question: How to change of the value an array element (in this case array[2]) in assembly?
assembly gas att
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
I have the following C code that returns 9:
#include <stdio.h>
int main()
{
int array[4]={0};
array[2]=9;
return array[2];
}
To accomplish the same in assembly language I have tried:
.section .data
array: .zero 4
.section .text
.globl _start
_start:
mov $2,%esi
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx
mov $9,%ecx
mov $1,%eax
mov array(,%esi,4),%ebx # return the value of array[2]
int $0x80
assembled and linked with:
gcc -g -c array.s && ld array.o && ./a.out
However this program returns 0 instead of 9:
>./a.out
>echo $?
0
I suspect the problem is in the lines with comments. After various unsuccessful attempts to find the problem I decided to ask the question: How to change of the value an array element (in this case array[2]) in assembly?
assembly gas att
assembly gas att
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
edited Nov 23 '18 at 15:49
Sergio
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
asked Nov 23 '18 at 15:43
SergioSergio
3,3563927
3,3563927
2
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx- this loads value from memory. To get address in similar way one can uselea array(,%esi,4),%ecx. Andmov $9,%ecxloads value 9 intoecx, it does not involve any extra memory access except the instruction opcode itself. And your finalmov array(,%esi,4),%ebxis completely correct and doing what it should, but if you will check how it differs from that "move address to ecx", there's no difference, so I'm not sure why you expect different result. (also using debugger to watch CPU state per instruction is of tremendous help)
– Ped7g
Nov 23 '18 at 15:50
1
And to modify array in fixed way, you can skip all those extra preparations, and just do immediatelymovl $9, array+2*4to use theMOV r/m32, imm32instruction to write immediate directly into memory. If you insist on indexing (to learn it for later), thenmovl $9, array(,%esi,4)writes value 9 into memory at "array + esi*4". To use "address" inecx(if it would hold one), you can domovl $9,(%ecx), to write into memory, the parentheses make it memory operand, not register, as target.
– Ped7g
Nov 23 '18 at 15:54
Thanks, now I see the flaws with the code I posted. Bothmovl $9, array+2*4andmovl $9,array(,%esi,4)give the desired result.
– Sergio
Nov 23 '18 at 16:46
add a comment |
2
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx- this loads value from memory. To get address in similar way one can uselea array(,%esi,4),%ecx. Andmov $9,%ecxloads value 9 intoecx, it does not involve any extra memory access except the instruction opcode itself. And your finalmov array(,%esi,4),%ebxis completely correct and doing what it should, but if you will check how it differs from that "move address to ecx", there's no difference, so I'm not sure why you expect different result. (also using debugger to watch CPU state per instruction is of tremendous help)
– Ped7g
Nov 23 '18 at 15:50
1
And to modify array in fixed way, you can skip all those extra preparations, and just do immediatelymovl $9, array+2*4to use theMOV r/m32, imm32instruction to write immediate directly into memory. If you insist on indexing (to learn it for later), thenmovl $9, array(,%esi,4)writes value 9 into memory at "array + esi*4". To use "address" inecx(if it would hold one), you can domovl $9,(%ecx), to write into memory, the parentheses make it memory operand, not register, as target.
– Ped7g
Nov 23 '18 at 15:54
Thanks, now I see the flaws with the code I posted. Bothmovl $9, array+2*4andmovl $9,array(,%esi,4)give the desired result.
– Sergio
Nov 23 '18 at 16:46
2
2
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx - this loads value from memory. To get address in similar way one can use lea array(,%esi,4),%ecx. And mov $9,%ecx loads value 9 into ecx, it does not involve any extra memory access except the instruction opcode itself. And your final mov array(,%esi,4),%ebx is completely correct and doing what it should, but if you will check how it differs from that "move address to ecx", there's no difference, so I'm not sure why you expect different result. (also using debugger to watch CPU state per instruction is of tremendous help)– Ped7g
Nov 23 '18 at 15:50
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx - this loads value from memory. To get address in similar way one can use lea array(,%esi,4),%ecx. And mov $9,%ecx loads value 9 into ecx, it does not involve any extra memory access except the instruction opcode itself. And your final mov array(,%esi,4),%ebx is completely correct and doing what it should, but if you will check how it differs from that "move address to ecx", there's no difference, so I'm not sure why you expect different result. (also using debugger to watch CPU state per instruction is of tremendous help)– Ped7g
Nov 23 '18 at 15:50
1
1
And to modify array in fixed way, you can skip all those extra preparations, and just do immediately
movl $9, array+2*4 to use the MOV r/m32, imm32 instruction to write immediate directly into memory. If you insist on indexing (to learn it for later), then movl $9, array(,%esi,4) writes value 9 into memory at "array + esi*4". To use "address" in ecx (if it would hold one), you can do movl $9,(%ecx), to write into memory, the parentheses make it memory operand, not register, as target.– Ped7g
Nov 23 '18 at 15:54
And to modify array in fixed way, you can skip all those extra preparations, and just do immediately
movl $9, array+2*4 to use the MOV r/m32, imm32 instruction to write immediate directly into memory. If you insist on indexing (to learn it for later), then movl $9, array(,%esi,4) writes value 9 into memory at "array + esi*4". To use "address" in ecx (if it would hold one), you can do movl $9,(%ecx), to write into memory, the parentheses make it memory operand, not register, as target.– Ped7g
Nov 23 '18 at 15:54
Thanks, now I see the flaws with the code I posted. Both
movl $9, array+2*4 and movl $9,array(,%esi,4) give the desired result.– Sergio
Nov 23 '18 at 16:46
Thanks, now I see the flaws with the code I posted. Both
movl $9, array+2*4 and movl $9,array(,%esi,4) give the desired result.– Sergio
Nov 23 '18 at 16:46
add a comment |
1 Answer
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.section .data
array: .zero 16 # allocate 16 bytes instead of 4
.section .text
.globl _start
_start:
mov $2,%esi
lea array(,%esi,4),%ecx # copy address of array[2] to %ecx, not the value
mov $9, %ebx
mov %ebx,(%ecx) # assign value into the memory pointed by %ecx
mov $1,%eax
mov array(,%esi,4),%ebx
int $0x80
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
The linemov $9,(%ecx)gives the error:no instruction mnemonic suffix given and no register operands;
– Sergio
Nov 23 '18 at 16:44
Put the value in a register (%ebx) before moving it in memory, or just addlsuffix to the mnemonic:movl $9, (%ecx)
– Nathan Xabedi
Nov 23 '18 at 16:57
1
mov $9,(%ecx)is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions likemovl / movw / movb /..., but as long as there's no ambiguity, thegaswill accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you writemovw %eax,(%ebp)=eaxerror.
– Ped7g
Nov 23 '18 at 22:12
add a comment |
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.section .data
array: .zero 16 # allocate 16 bytes instead of 4
.section .text
.globl _start
_start:
mov $2,%esi
lea array(,%esi,4),%ecx # copy address of array[2] to %ecx, not the value
mov $9, %ebx
mov %ebx,(%ecx) # assign value into the memory pointed by %ecx
mov $1,%eax
mov array(,%esi,4),%ebx
int $0x80
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
The linemov $9,(%ecx)gives the error:no instruction mnemonic suffix given and no register operands;
– Sergio
Nov 23 '18 at 16:44
Put the value in a register (%ebx) before moving it in memory, or just addlsuffix to the mnemonic:movl $9, (%ecx)
– Nathan Xabedi
Nov 23 '18 at 16:57
1
mov $9,(%ecx)is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions likemovl / movw / movb /..., but as long as there's no ambiguity, thegaswill accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you writemovw %eax,(%ebp)=eaxerror.
– Ped7g
Nov 23 '18 at 22:12
add a comment |
0
.section .data
array: .zero 16 # allocate 16 bytes instead of 4
.section .text
.globl _start
_start:
mov $2,%esi
lea array(,%esi,4),%ecx # copy address of array[2] to %ecx, not the value
mov $9, %ebx
mov %ebx,(%ecx) # assign value into the memory pointed by %ecx
mov $1,%eax
mov array(,%esi,4),%ebx
int $0x80
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
The linemov $9,(%ecx)gives the error:no instruction mnemonic suffix given and no register operands;
– Sergio
Nov 23 '18 at 16:44
Put the value in a register (%ebx) before moving it in memory, or just addlsuffix to the mnemonic:movl $9, (%ecx)
– Nathan Xabedi
Nov 23 '18 at 16:57
1
mov $9,(%ecx)is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions likemovl / movw / movb /..., but as long as there's no ambiguity, thegaswill accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you writemovw %eax,(%ebp)=eaxerror.
– Ped7g
Nov 23 '18 at 22:12
add a comment |
0
0
0
.section .data
array: .zero 16 # allocate 16 bytes instead of 4
.section .text
.globl _start
_start:
mov $2,%esi
lea array(,%esi,4),%ecx # copy address of array[2] to %ecx, not the value
mov $9, %ebx
mov %ebx,(%ecx) # assign value into the memory pointed by %ecx
mov $1,%eax
mov array(,%esi,4),%ebx
int $0x80
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
.section .data
array: .zero 16 # allocate 16 bytes instead of 4
.section .text
.globl _start
_start:
mov $2,%esi
lea array(,%esi,4),%ecx # copy address of array[2] to %ecx, not the value
mov $9, %ebx
mov %ebx,(%ecx) # assign value into the memory pointed by %ecx
mov $1,%eax
mov array(,%esi,4),%ebx
int $0x80
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
edited Nov 23 '18 at 16:55
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
answered Nov 23 '18 at 16:26
Nathan XabediNathan Xabedi
1036
1036
The linemov $9,(%ecx)gives the error:no instruction mnemonic suffix given and no register operands;
– Sergio
Nov 23 '18 at 16:44
Put the value in a register (%ebx) before moving it in memory, or just addlsuffix to the mnemonic:movl $9, (%ecx)
– Nathan Xabedi
Nov 23 '18 at 16:57
1
mov $9,(%ecx)is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions likemovl / movw / movb /..., but as long as there's no ambiguity, thegaswill accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you writemovw %eax,(%ebp)=eaxerror.
– Ped7g
Nov 23 '18 at 22:12
add a comment |
The linemov $9,(%ecx)gives the error:no instruction mnemonic suffix given and no register operands;
– Sergio
Nov 23 '18 at 16:44
Put the value in a register (%ebx) before moving it in memory, or just addlsuffix to the mnemonic:movl $9, (%ecx)
– Nathan Xabedi
Nov 23 '18 at 16:57
1
mov $9,(%ecx)is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions likemovl / movw / movb /..., but as long as there's no ambiguity, thegaswill accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you writemovw %eax,(%ebp)=eaxerror.
– Ped7g
Nov 23 '18 at 22:12
The line
mov $9,(%ecx) gives the error: no instruction mnemonic suffix given and no register operands;– Sergio
Nov 23 '18 at 16:44
The line
mov $9,(%ecx) gives the error: no instruction mnemonic suffix given and no register operands;– Sergio
Nov 23 '18 at 16:44
Put the value in a register (
%ebx) before moving it in memory, or just add l suffix to the mnemonic: movl $9, (%ecx)– Nathan Xabedi
Nov 23 '18 at 16:57
Put the value in a register (
%ebx) before moving it in memory, or just add l suffix to the mnemonic: movl $9, (%ecx)– Nathan Xabedi
Nov 23 '18 at 16:57
1
1
mov $9,(%ecx) is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions like movl / movw / movb /..., but as long as there's no ambiguity, the gas will accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you write movw %eax,(%ebp) = eax error.– Ped7g
Nov 23 '18 at 22:12
mov $9,(%ecx) is ambiguous, the assembler doesn't know if byte, word or dword should be written in memory (all three options are possible in x86 32 bit mode). The AT&T syntax is mostly used with suffix form of instructions like movl / movw / movb /..., but as long as there's no ambiguity, the gas will accept also missing suffix, like OP original code has everywhere. Writing suffix form everywhere is a bit more writing (and I personally have harder time to read it), but it may help to catch some bugs, like you want to store word into memory, and you write movw %eax,(%ebp) = eax error.– Ped7g
Nov 23 '18 at 22:12
add a comment |
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2
mov array(,%esi,4),%ecx # copy address of array[2] to %ecx- this loads value from memory. To get address in similar way one can uselea array(,%esi,4),%ecx. Andmov $9,%ecxloads value 9 intoecx, it does not involve any extra memory access except the instruction opcode itself. And your finalmov array(,%esi,4),%ebxis completely correct and doing what it should, but if you will check how it differs from that "move address to ecx", there's no difference, so I'm not sure why you expect different result. (also using debugger to watch CPU state per instruction is of tremendous help)– Ped7g
Nov 23 '18 at 15:50
1
And to modify array in fixed way, you can skip all those extra preparations, and just do immediately
movl $9, array+2*4to use theMOV r/m32, imm32instruction to write immediate directly into memory. If you insist on indexing (to learn it for later), thenmovl $9, array(,%esi,4)writes value 9 into memory at "array + esi*4". To use "address" inecx(if it would hold one), you can domovl $9,(%ecx), to write into memory, the parentheses make it memory operand, not register, as target.– Ped7g
Nov 23 '18 at 15:54
Thanks, now I see the flaws with the code I posted. Both
movl $9, array+2*4andmovl $9,array(,%esi,4)give the desired result.– Sergio
Nov 23 '18 at 16:46