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I want to swap the position of two bits. How to do this?

The Next CEO of Stack OverflowWhat are the segment and offset in real mode memory addressing?How to right rotate a 64-bit value efficiently in ARM7 assembler?IA-32e 64-bit IDT Gate DescriptorBase pointer and stack pointerx86 - segmentation in protected mode serves what purpose?conceptual help understanding OFFSET Function and the ways in which this program outputsReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsAccessing the Program Segment PrefixWhat happens to instruction pointers when address overrides are used to target a smaller address space?Verify that in a range of bits only one is on

I am given a question in which I have to write a program such that the last bit of the offset address of the first segment becomes the first bit of the offset address of the second segment. For example if I am given ABCDH then the offset address of the second address should be DCBAH. I am just focusing on the swapping of the offset address and am ignoring the base address for now:

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH. Now I want to swap the position of D and C. I am stuck at this point. Will I use SAR command?

edited Mar 10 at 14:38

asked Mar 7 at 15:22

Ahmad Qayyum

335

5

You have mixed 1, 2 and 4 bit units all over the place. If you really need to turn ABCDh into DCBAh that's reversing using 4 bit units. Obviously ror ax, 2 will rotate by 2 bits which will not give CDABh as you claim. To get CDABh you want to rotate by 2 bytes which is 16 bits.

– Jester
Mar 7 at 15:49

2

@Jester: 2 hex digits = 1 byte, so ror ax, 8. If C, D, A, B are placeholders for whole bytes, then yes ror eax, 16. This question is so mixed up between bits, nibbles, and bytes that it apparently got you mixed up :P

– Peter Cordes
Mar 7 at 23:07

Reversing the bits of 0ABCDH (not ABCDH, by the way) results in 0B3D5H, not 0DCBAH.

– TonyK
Mar 10 at 17:47

add a comment |

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH. Now I want to swap the position of D and C. I am stuck at this point. Will I use SAR command?

edited Mar 10 at 14:38

asked Mar 7 at 15:22

Ahmad Qayyum

335

5

You have mixed 1, 2 and 4 bit units all over the place. If you really need to turn ABCDh into DCBAh that's reversing using 4 bit units. Obviously ror ax, 2 will rotate by 2 bits which will not give CDABh as you claim. To get CDABh you want to rotate by 2 bytes which is 16 bits.

– Jester
Mar 7 at 15:49

2

@Jester: 2 hex digits = 1 byte, so ror ax, 8. If C, D, A, B are placeholders for whole bytes, then yes ror eax, 16. This question is so mixed up between bits, nibbles, and bytes that it apparently got you mixed up :P

– Peter Cordes
Mar 7 at 23:07

Reversing the bits of 0ABCDH (not ABCDH, by the way) results in 0B3D5H, not 0DCBAH.

– TonyK
Mar 10 at 17:47

add a comment |

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH. Now I want to swap the position of D and C. I am stuck at this point. Will I use SAR command?

edited Mar 10 at 14:38

asked Mar 7 at 15:22

Ahmad Qayyum

335

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH. Now I want to swap the position of D and C. I am stuck at this point. Will I use SAR command?

assembly x86 x86-emulation

edited Mar 10 at 14:38

asked Mar 7 at 15:22

Ahmad Qayyum

335

edited Mar 10 at 14:38

asked Mar 7 at 15:22

Ahmad Qayyum

335

edited Mar 10 at 14:38

asked Mar 7 at 15:22

Ahmad Qayyum

335

asked Mar 7 at 15:22

Ahmad Qayyum

335

asked Mar 7 at 15:22

Ahmad Qayyum

335

5

You have mixed 1, 2 and 4 bit units all over the place. If you really need to turn ABCDh into DCBAh that's reversing using 4 bit units. Obviously ror ax, 2 will rotate by 2 bits which will not give CDABh as you claim. To get CDABh you want to rotate by 2 bytes which is 16 bits.

– Jester
Mar 7 at 15:49

2

@Jester: 2 hex digits = 1 byte, so ror ax, 8. If C, D, A, B are placeholders for whole bytes, then yes ror eax, 16. This question is so mixed up between bits, nibbles, and bytes that it apparently got you mixed up :P

– Peter Cordes
Mar 7 at 23:07

Reversing the bits of 0ABCDH (not ABCDH, by the way) results in 0B3D5H, not 0DCBAH.

– TonyK
Mar 10 at 17:47

add a comment |

5

You have mixed 1, 2 and 4 bit units all over the place. If you really need to turn ABCDh into DCBAh that's reversing using 4 bit units. Obviously ror ax, 2 will rotate by 2 bits which will not give CDABh as you claim. To get CDABh you want to rotate by 2 bytes which is 16 bits.

– Jester
Mar 7 at 15:49

2

@Jester: 2 hex digits = 1 byte, so ror ax, 8. If C, D, A, B are placeholders for whole bytes, then yes ror eax, 16. This question is so mixed up between bits, nibbles, and bytes that it apparently got you mixed up :P

– Peter Cordes
Mar 7 at 23:07

Reversing the bits of 0ABCDH (not ABCDH, by the way) results in 0B3D5H, not 0DCBAH.

– TonyK
Mar 10 at 17:47

You have mixed 1, 2 and 4 bit units all over the place. If you really need to turn ABCDh into DCBAh that's reversing using 4 bit units. Obviously ror ax, 2 will rotate by 2 bits which will not give CDABh as you claim. To get CDABh you want to rotate by 2 bytes which is 16 bits.

– Jester
Mar 7 at 15:49

@Jester: 2 hex digits = 1 byte, so ror ax, 8. If C, D, A, B are placeholders for whole bytes, then yes ror eax, 16. This question is so mixed up between bits, nibbles, and bytes that it apparently got you mixed up :P

– Peter Cordes
Mar 7 at 23:07

Reversing the bits of 0ABCDH (not ABCDH, by the way) results in 0B3D5H, not 0DCBAH.

– TonyK
Mar 10 at 17:47

add a comment |

1 Answer
1

active

oldest

votes

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH

The AX register holds 16 bits. When you rotate these 16 bits 16 times you get the same value that you started with!

For example if I am given ABCDH then the offset address of the second address should be DCBAH

So you want to go from ABCDh to DCBAh.

The AX register is subdivided in 2 halves. The low half is named AL and the high half is named AH. You can operate on these halves independently.

The instruction mov ax, 0ABCDh puts the value 0ABh in AH and puts the value 0CDh in AL.

mov ax, 0ABCDh ; AH = 0ABh AL = 0CDh
rol al, 4 ; AH = 0ABh AL = 0DCh 
rol ah, 4 ; AH = 0BAh AL = 0DCh
xchg al, ah ; AH = 0DCh AL = 0BAh

Now finally AX=0DCBAh.

All of the above deals with 4-bit quantities. We call these nibbles.

You can write your hexadecimal value 0ABCDh using the binary representation like 1010101111001101b. You can see that there are 16 bits.

Aligned groups of bits have special names:

every 4 bits form a nibble, you can see that there are 4 nibbles. (1010 1011 1100 1101)

every 8 bits form a byte, you can see that there are 2 bytes. (10101011 11001101)

every 16 bits form a word, you can see that there is 1 word. (1010101111001101)

edited Mar 14 at 15:15

Fifoernik

7,79511324

answered Mar 10 at 20:50

Sep Roland

12.4k22047

add a comment |

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MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH

The AX register holds 16 bits. When you rotate these 16 bits 16 times you get the same value that you started with!

For example if I am given ABCDH then the offset address of the second address should be DCBAH

So you want to go from ABCDh to DCBAh.

The AX register is subdivided in 2 halves. The low half is named AL and the high half is named AH. You can operate on these halves independently.

The instruction mov ax, 0ABCDh puts the value 0ABh in AH and puts the value 0CDh in AL.

mov ax, 0ABCDh ; AH = 0ABh AL = 0CDh
rol al, 4 ; AH = 0ABh AL = 0DCh 
rol ah, 4 ; AH = 0BAh AL = 0DCh
xchg al, ah ; AH = 0DCh AL = 0BAh

Now finally AX=0DCBAh.

Aligned groups of bits have special names:

every 4 bits form a nibble, you can see that there are 4 nibbles. (1010 1011 1100 1101)

every 8 bits form a byte, you can see that there are 2 bytes. (10101011 11001101)

every 16 bits form a word, you can see that there is 1 word. (1010101111001101)

edited Mar 14 at 15:15

Fifoernik

7,79511324

answered Mar 10 at 20:50

Sep Roland

12.4k22047

add a comment |

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH

The AX register holds 16 bits. When you rotate these 16 bits 16 times you get the same value that you started with!

For example if I am given ABCDH then the offset address of the second address should be DCBAH

So you want to go from ABCDh to DCBAh.

The AX register is subdivided in 2 halves. The low half is named AL and the high half is named AH. You can operate on these halves independently.

The instruction mov ax, 0ABCDh puts the value 0ABh in AH and puts the value 0CDh in AL.

mov ax, 0ABCDh ; AH = 0ABh AL = 0CDh
rol al, 4 ; AH = 0ABh AL = 0DCh 
rol ah, 4 ; AH = 0BAh AL = 0DCh
xchg al, ah ; AH = 0DCh AL = 0BAh

Now finally AX=0DCBAh.

Aligned groups of bits have special names:

every 4 bits form a nibble, you can see that there are 4 nibbles. (1010 1011 1100 1101)

every 8 bits form a byte, you can see that there are 2 bytes. (10101011 11001101)

every 16 bits form a word, you can see that there is 1 word. (1010101111001101)

edited Mar 14 at 15:15

Fifoernik

7,79511324

answered Mar 10 at 20:50

Sep Roland

12.4k22047

add a comment |

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH

The AX register holds 16 bits. When you rotate these 16 bits 16 times you get the same value that you started with!

For example if I am given ABCDH then the offset address of the second address should be DCBAH

So you want to go from ABCDh to DCBAh.

The AX register is subdivided in 2 halves. The low half is named AL and the high half is named AH. You can operate on these halves independently.

The instruction mov ax, 0ABCDh puts the value 0ABh in AH and puts the value 0CDh in AL.

mov ax, 0ABCDh ; AH = 0ABh AL = 0CDh
rol al, 4 ; AH = 0ABh AL = 0DCh 
rol ah, 4 ; AH = 0BAh AL = 0DCh
xchg al, ah ; AH = 0DCh AL = 0BAh

Now finally AX=0DCBAh.

Aligned groups of bits have special names:

every 4 bits form a nibble, you can see that there are 4 nibbles. (1010 1011 1100 1101)

every 8 bits form a byte, you can see that there are 2 bytes. (10101011 11001101)

every 16 bits form a word, you can see that there is 1 word. (1010101111001101)

edited Mar 14 at 15:15

Fifoernik

7,79511324

answered Mar 10 at 20:50

Sep Roland

12.4k22047

MOV AX,ABCDH
ROR AX,16 ; this will rotate the value of AX 16 times

Now we have CDABH

The AX register holds 16 bits. When you rotate these 16 bits 16 times you get the same value that you started with!

For example if I am given ABCDH then the offset address of the second address should be DCBAH

So you want to go from ABCDh to DCBAh.

The AX register is subdivided in 2 halves. The low half is named AL and the high half is named AH. You can operate on these halves independently.

The instruction mov ax, 0ABCDh puts the value 0ABh in AH and puts the value 0CDh in AL.

mov ax, 0ABCDh ; AH = 0ABh AL = 0CDh
rol al, 4 ; AH = 0ABh AL = 0DCh 
rol ah, 4 ; AH = 0BAh AL = 0DCh
xchg al, ah ; AH = 0DCh AL = 0BAh

Now finally AX=0DCBAh.

Aligned groups of bits have special names:

every 4 bits form a nibble, you can see that there are 4 nibbles. (1010 1011 1100 1101)

every 8 bits form a byte, you can see that there are 2 bytes. (10101011 11001101)

every 16 bits form a word, you can see that there is 1 word. (1010101111001101)

edited Mar 14 at 15:15

Fifoernik

7,79511324

answered Mar 10 at 20:50

Sep Roland

12.4k22047

edited Mar 14 at 15:15

Fifoernik

7,79511324

edited Mar 14 at 15:15

Fifoernik

7,79511324

edited Mar 14 at 15:15

Fifoernik

7,79511324

answered Mar 10 at 20:50

Sep Roland

12.4k22047

answered Mar 10 at 20:50

Sep Roland

12.4k22047

answered Mar 10 at 20:50

Sep Roland

12.4k22047

add a comment |

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