Mixing
ARM assembly question wiht hexidecimal values
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 '18 at 23:56
MangoKittyMangoKitty
223
add a comment |
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 '18 at 23:56
MangoKittyMangoKitty
223
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 '18 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 '18 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 '18 at 14:55
add a comment |
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 '18 at 23:56
MangoKittyMangoKitty
223
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 '18 at 23:56
MangoKittyMangoKitty
223
asked Nov 21 '18 at 23:56
MangoKittyMangoKitty
223
asked Nov 21 '18 at 23:56
MangoKittyMangoKitty
223
asked Nov 21 '18 at 23:56
MangoKittyMangoKitty
223
asked Nov 21 '18 at 23:56
MangoKittyMangoKitty
223
223
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 '18 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 '18 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 '18 at 14:55
add a comment |
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 '18 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 '18 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 '18 at 14:55
4
4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 '18 at 0:11
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 '18 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 '18 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 '18 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 '18 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 '18 at 14:55
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 0:51
DavidPMDavidPM
33518
add a comment |
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 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
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 '18 at 0:51
DavidPMDavidPM
33518
add a comment |
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 '18 at 0:51
DavidPMDavidPM
33518
add a comment |
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 '18 at 0:51
DavidPMDavidPM
33518
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 '18 at 0:51
DavidPMDavidPM
33518
answered Nov 22 '18 at 0:51
DavidPMDavidPM
33518
answered Nov 22 '18 at 0:51
DavidPMDavidPM
33518
answered Nov 22 '18 at 0:51
DavidPMDavidPM
33518
33518
add a comment |
add a comment |
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 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
add a comment |
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 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
add a comment |
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 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
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 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
answered Nov 22 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
answered Nov 22 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
answered Nov 22 '18 at 9:43
Yves DaoustYves Daoust
37.7k72659
37.7k72659
add a comment |
add a comment |
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4
yes. The leading zeros contribute nothing to the real value
– phuclv
Nov 22 '18 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 '18 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 '18 at 14:55