ARM assembly question wiht hexidecimal values
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If i have 0x00000065 stored in a register, is that the same as having 0X65 in my register?
Thank you so much.
math assembly arm hex
asked Nov 21 at 23:56
MangoKitty
162
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up vote
0
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favorite
If i have 0x00000065 stored in a register, is that the same as having 0X65 in my register?
Thank you so much.
math assembly arm hex
asked Nov 21 at 23:56
MangoKitty
162
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 at 0:11
1
It's the same with the denary numbers you're used to, if you have $2 or $0000002 you have the same amount of money.
– Colin
Nov 22 at 7:55
what does this have to do with assembly, and if it is an assembly question show the assembly and indicate which assembler you are using as the assembler determines how numbers are interpreted/used not the target instruction set.
– old_timer
Nov 22 at 14:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If i have 0x00000065 stored in a register, is that the same as having 0X65 in my register?
Thank you so much.
math assembly arm hex
asked Nov 21 at 23:56
MangoKitty
162
If i have 0x00000065 stored in a register, is that the same as having 0X65 in my register?
Thank you so much.
math assembly arm hex
math assembly arm hex
asked Nov 21 at 23:56
MangoKitty
162
asked Nov 21 at 23:56
MangoKitty
162
asked Nov 21 at 23:56
MangoKitty
162
asked Nov 21 at 23:56
MangoKitty
162
asked Nov 21 at 23:56
MangoKitty
162
162
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 at 0:11
1
It's the same with the denary numbers you're used to, if you have $2 or $0000002 you have the same amount of money.
– Colin
Nov 22 at 7:55
what does this have to do with assembly, and if it is an assembly question show the assembly and indicate which assembler you are using as the assembler determines how numbers are interpreted/used not the target instruction set.
– old_timer
Nov 22 at 14:55
add a comment |
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 at 0:11
1
It's the same with the denary numbers you're used to, if you have $2 or $0000002 you have the same amount of money.
– Colin
Nov 22 at 7:55
what does this have to do with assembly, and if it is an assembly question show the assembly and indicate which assembler you are using as the assembler determines how numbers are interpreted/used not the target instruction set.
– old_timer
Nov 22 at 14:55
4
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 at 0:11
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 at 0:11
1
1
It's the same with the denary numbers you're used to, if you have $2 or $0000002 you have the same amount of money.
– Colin
Nov 22 at 7:55
It's the same with the denary numbers you're used to, if you have $2 or $0000002 you have the same amount of money.
– Colin
Nov 22 at 7:55
what does this have to do with assembly, and if it is an assembly question show the assembly and indicate which assembler you are using as the assembler determines how numbers are interpreted/used not the target instruction set.
– old_timer
Nov 22 at 14:55
what does this have to do with assembly, and if it is an assembly question show the assembly and indicate which assembler you are using as the assembler determines how numbers are interpreted/used not the target instruction set.
– old_timer
Nov 22 at 14:55
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Yes, it's the same two hexadecimal values:
0x00000065 = 5*(16^0) + 6*(16^1) + 0*(16^2) + ... + 0*(16^7) = 5*(16^0) + 6*(16^1) = 0x65
(Note: the symbol '^' denotes the power operator)
answered Nov 22 at 0:51
DavidPM
915
add a comment |
up vote
-3
down vote
Registers are 32 bits long so you can't have 0x65 in one, only 0x00000065.
But of course, these are equal numbers.
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Yes, it's the same two hexadecimal values:
0x00000065 = 5*(16^0) + 6*(16^1) + 0*(16^2) + ... + 0*(16^7) = 5*(16^0) + 6*(16^1) = 0x65
(Note: the symbol '^' denotes the power operator)
answered Nov 22 at 0:51
DavidPM
915
add a comment |
up vote
1
down vote
Yes, it's the same two hexadecimal values:
0x00000065 = 5*(16^0) + 6*(16^1) + 0*(16^2) + ... + 0*(16^7) = 5*(16^0) + 6*(16^1) = 0x65
(Note: the symbol '^' denotes the power operator)
answered Nov 22 at 0:51
DavidPM
915
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes, it's the same two hexadecimal values:
0x00000065 = 5*(16^0) + 6*(16^1) + 0*(16^2) + ... + 0*(16^7) = 5*(16^0) + 6*(16^1) = 0x65
(Note: the symbol '^' denotes the power operator)
answered Nov 22 at 0:51
DavidPM
915
Yes, it's the same two hexadecimal values:
0x00000065 = 5*(16^0) + 6*(16^1) + 0*(16^2) + ... + 0*(16^7) = 5*(16^0) + 6*(16^1) = 0x65
(Note: the symbol '^' denotes the power operator)
answered Nov 22 at 0:51
DavidPM
915
answered Nov 22 at 0:51
DavidPM
915
answered Nov 22 at 0:51
DavidPM
915
answered Nov 22 at 0:51
DavidPM
915
915
add a comment |
add a comment |
up vote
-3
down vote
Registers are 32 bits long so you can't have 0x65 in one, only 0x00000065.
But of course, these are equal numbers.
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
add a comment |
up vote
-3
down vote
Registers are 32 bits long so you can't have 0x65 in one, only 0x00000065.
But of course, these are equal numbers.
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
add a comment |
up vote
-3
down vote
up vote
-3
down vote
Registers are 32 bits long so you can't have 0x65 in one, only 0x00000065.
But of course, these are equal numbers.
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
Registers are 32 bits long so you can't have 0x65 in one, only 0x00000065.
But of course, these are equal numbers.
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
answered Nov 22 at 9:43
Yves Daoust
36.2k72559
36.2k72559
add a comment |
add a comment |
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4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 at 0:11
1
It's the same with the denary numbers you're used to, if you have $2 or $0000002 you have the same amount of money.
– Colin
Nov 22 at 7:55
what does this have to do with assembly, and if it is an assembly question show the assembly and indicate which assembler you are using as the assembler determines how numbers are interpreted/used not the target instruction set.
– old_timer
Nov 22 at 14:55