Risc-v instruction that i dont understand

I have this risc v code :

lui S0, 0x1234
ori S1, S0, 0x5678
add S2, S1, S1

and the question asks me, "What does the register S2 hold?"
The question explains that lui and I quote:

"Load the lower half word of the immediate imm into the upper halfword of register rt. The lower bits of the register are set to 0"

I don't know how to 'compile this program' and what does 0x1234 mean? Thanks

edited Nov 14 '18 at 19:29

dopamane

85611024

asked Nov 14 '18 at 17:37

Razi Awad

0x1234 is a hexadecimal number equal to 4660 (decimal).

– Amy
Nov 14 '18 at 19:31

ok and what does the lui do

– Razi Awad
Nov 14 '18 at 20:08

You state what it does in the question.

– Amy
Nov 14 '18 at 21:23

stackoverflow.com/q/50742420/995714

– phuclv
Nov 15 '18 at 3:14

Error: illegal operands ori s1,s0,0x5678 -- immediate operand is 12 bits only (sign extended)

– Pavel Smirnov
Nov 23 '18 at 21:24

add a comment |

I have this risc v code :

lui S0, 0x1234
ori S1, S0, 0x5678
add S2, S1, S1

and the question asks me, "What does the register S2 hold?"
The question explains that lui and I quote:

"Load the lower half word of the immediate imm into the upper halfword of register rt. The lower bits of the register are set to 0"

I don't know how to 'compile this program' and what does 0x1234 mean? Thanks

edited Nov 14 '18 at 19:29

dopamane

85611024

asked Nov 14 '18 at 17:37

Razi Awad

0x1234 is a hexadecimal number equal to 4660 (decimal).

– Amy
Nov 14 '18 at 19:31

ok and what does the lui do

– Razi Awad
Nov 14 '18 at 20:08

You state what it does in the question.

– Amy
Nov 14 '18 at 21:23

stackoverflow.com/q/50742420/995714

– phuclv
Nov 15 '18 at 3:14

Error: illegal operands ori s1,s0,0x5678 -- immediate operand is 12 bits only (sign extended)

– Pavel Smirnov
Nov 23 '18 at 21:24

add a comment |

I have this risc v code :

lui S0, 0x1234
ori S1, S0, 0x5678
add S2, S1, S1

and the question asks me, "What does the register S2 hold?"
The question explains that lui and I quote:

"Load the lower half word of the immediate imm into the upper halfword of register rt. The lower bits of the register are set to 0"

I don't know how to 'compile this program' and what does 0x1234 mean? Thanks

edited Nov 14 '18 at 19:29

dopamane

85611024

asked Nov 14 '18 at 17:37

Razi Awad

I have this risc v code :

lui S0, 0x1234
ori S1, S0, 0x5678
add S2, S1, S1

and the question asks me, "What does the register S2 hold?"
The question explains that lui and I quote:

"Load the lower half word of the immediate imm into the upper halfword of register rt. The lower bits of the register are set to 0"

I don't know how to 'compile this program' and what does 0x1234 mean? Thanks

instruction-set riscv

edited Nov 14 '18 at 19:29

dopamane

85611024

asked Nov 14 '18 at 17:37

Razi Awad

edited Nov 14 '18 at 19:29

dopamane

85611024

asked Nov 14 '18 at 17:37

Razi Awad

edited Nov 14 '18 at 19:29

dopamane

85611024

edited Nov 14 '18 at 19:29

dopamane

85611024

edited Nov 14 '18 at 19:29

dopamane

85611024

asked Nov 14 '18 at 17:37

Razi Awad

asked Nov 14 '18 at 17:37

Razi Awad

asked Nov 14 '18 at 17:37

Razi Awad

0x1234 is a hexadecimal number equal to 4660 (decimal).

– Amy
Nov 14 '18 at 19:31

ok and what does the lui do

– Razi Awad
Nov 14 '18 at 20:08

You state what it does in the question.

– Amy
Nov 14 '18 at 21:23

stackoverflow.com/q/50742420/995714

– phuclv
Nov 15 '18 at 3:14

Error: illegal operands ori s1,s0,0x5678 -- immediate operand is 12 bits only (sign extended)

– Pavel Smirnov
Nov 23 '18 at 21:24

add a comment |

0x1234 is a hexadecimal number equal to 4660 (decimal).

– Amy
Nov 14 '18 at 19:31

ok and what does the lui do

– Razi Awad
Nov 14 '18 at 20:08

You state what it does in the question.

– Amy
Nov 14 '18 at 21:23

stackoverflow.com/q/50742420/995714

– phuclv
Nov 15 '18 at 3:14

Error: illegal operands ori s1,s0,0x5678 -- immediate operand is 12 bits only (sign extended)

– Pavel Smirnov
Nov 23 '18 at 21:24

0x1234 is a hexadecimal number equal to 4660 (decimal).

– Amy
Nov 14 '18 at 19:31

ok and what does the lui do

– Razi Awad
Nov 14 '18 at 20:08

You state what it does in the question.

– Amy
Nov 14 '18 at 21:23

stackoverflow.com/q/50742420/995714

– phuclv
Nov 15 '18 at 3:14

Error: illegal operands ori s1,s0,0x5678 -- immediate operand is 12 bits only (sign extended)

– Pavel Smirnov
Nov 23 '18 at 21:24

add a comment |

2 Answers
2

active

oldest

votes

Take the instructions one at a time. First the load-upper-immediate, take the immediate (0x1234) and "load" it into the upper half of the S0 register and zero out the lower half:

lui S0, 0x1234 

S0 = 0x12340000

Next the or-immediate, we OR the value in S0 with the value 0x5678:

ori S1, S0, 0x5678

 0x12340000
OR 0x00005678
 ----------
 0x12345678 = S1

Finally the add, we are adding the value in S1 to itself or, equivalently, multiplying the value in S1 by 2:

add S2, S1, S1

 0x12345678
+ 0x12345678
 ----------
 0x2468ACF0 = S2

So the value in S2 register is 0x2468ACF0. Note, I am assuming 32-bit words. An immediate is like a constant, so lui is an instruction that places a constant into the upper half of a register. In combination with ori, you can load an entire word-immediate into a register.

edited Nov 15 '18 at 2:56

answered Nov 15 '18 at 2:50

dopamane

85611024

thank you so much sir , finally i understand how to compile thanks to you. very useful

– Razi Awad
Nov 15 '18 at 7:39

can i ask you one tiny question ? i have read that the lui instruction moves the 16 bits to the higher and the other 16 bits to the lower . how did you determine which side is higher and which is lower . also the 0x1234 does it contain 4 bits ? (1234) i have problem understanding this

– Razi Awad
Nov 15 '18 at 7:42

becuase we should have written it like that : 0x1234 0000 0000 0000 0000 0000 0000 0000 which is 32 bits half of it higher whiich is far left and half of it lower far right , Is it like that ?

– Razi Awad
Nov 15 '18 at 7:45

0x1234 is base 16 (or hexadecimal), so each "digit" is four bits. 0x1234 (base 16) = 0001_0010_0011_0100 (base 2). A register has 32-bits or 8 hex digits. The spec fig2.1 shows the most-significant-bit (XLEN-1) to the left-side. So upper is left, lower is right.

– dopamane
Nov 15 '18 at 7:52

Also, these instructions are technically "assembled" by an "assembler". Basically the Risc-V assembler takes the instructions (or mnemonics) and encodes them into binary words which are "fed" to the cpu. The instructions that an assembler understands make up an "assembly language". Compiling is a term used for higher level languages like C/C++.

– dopamane
Nov 15 '18 at 8:04

|
show 1 more comment

"LUI places the U-immediate value in the top 20 bits of the destination register rd, filling in the lowest 12 bits with zeros." -- riscv-spec-v2.2

so,
lui s0, 0x1234

s0 should be 0x01234000

edited Nov 23 '18 at 2:54

answered Nov 23 '18 at 2:48

jimmy li

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Take the instructions one at a time. First the load-upper-immediate, take the immediate (0x1234) and "load" it into the upper half of the S0 register and zero out the lower half:

lui S0, 0x1234 

S0 = 0x12340000

Next the or-immediate, we OR the value in S0 with the value 0x5678:

ori S1, S0, 0x5678

 0x12340000
OR 0x00005678
 ----------
 0x12345678 = S1

Finally the add, we are adding the value in S1 to itself or, equivalently, multiplying the value in S1 by 2:

add S2, S1, S1

 0x12345678
+ 0x12345678
 ----------
 0x2468ACF0 = S2

edited Nov 15 '18 at 2:56

answered Nov 15 '18 at 2:50

dopamane

85611024

thank you so much sir , finally i understand how to compile thanks to you. very useful

– Razi Awad
Nov 15 '18 at 7:39

can i ask you one tiny question ? i have read that the lui instruction moves the 16 bits to the higher and the other 16 bits to the lower . how did you determine which side is higher and which is lower . also the 0x1234 does it contain 4 bits ? (1234) i have problem understanding this

– Razi Awad
Nov 15 '18 at 7:42

becuase we should have written it like that : 0x1234 0000 0000 0000 0000 0000 0000 0000 which is 32 bits half of it higher whiich is far left and half of it lower far right , Is it like that ?

– Razi Awad
Nov 15 '18 at 7:45

0x1234 is base 16 (or hexadecimal), so each "digit" is four bits. 0x1234 (base 16) = 0001_0010_0011_0100 (base 2). A register has 32-bits or 8 hex digits. The spec fig2.1 shows the most-significant-bit (XLEN-1) to the left-side. So upper is left, lower is right.

– dopamane
Nov 15 '18 at 7:52

Also, these instructions are technically "assembled" by an "assembler". Basically the Risc-V assembler takes the instructions (or mnemonics) and encodes them into binary words which are "fed" to the cpu. The instructions that an assembler understands make up an "assembly language". Compiling is a term used for higher level languages like C/C++.

– dopamane
Nov 15 '18 at 8:04

|
show 1 more comment

Take the instructions one at a time. First the load-upper-immediate, take the immediate (0x1234) and "load" it into the upper half of the S0 register and zero out the lower half:

lui S0, 0x1234 

S0 = 0x12340000

Next the or-immediate, we OR the value in S0 with the value 0x5678:

ori S1, S0, 0x5678

 0x12340000
OR 0x00005678
 ----------
 0x12345678 = S1

Finally the add, we are adding the value in S1 to itself or, equivalently, multiplying the value in S1 by 2:

add S2, S1, S1

 0x12345678
+ 0x12345678
 ----------
 0x2468ACF0 = S2

edited Nov 15 '18 at 2:56

answered Nov 15 '18 at 2:50

dopamane

85611024

thank you so much sir , finally i understand how to compile thanks to you. very useful

– Razi Awad
Nov 15 '18 at 7:39

can i ask you one tiny question ? i have read that the lui instruction moves the 16 bits to the higher and the other 16 bits to the lower . how did you determine which side is higher and which is lower . also the 0x1234 does it contain 4 bits ? (1234) i have problem understanding this

– Razi Awad
Nov 15 '18 at 7:42

becuase we should have written it like that : 0x1234 0000 0000 0000 0000 0000 0000 0000 which is 32 bits half of it higher whiich is far left and half of it lower far right , Is it like that ?

– Razi Awad
Nov 15 '18 at 7:45

0x1234 is base 16 (or hexadecimal), so each "digit" is four bits. 0x1234 (base 16) = 0001_0010_0011_0100 (base 2). A register has 32-bits or 8 hex digits. The spec fig2.1 shows the most-significant-bit (XLEN-1) to the left-side. So upper is left, lower is right.

– dopamane
Nov 15 '18 at 7:52

Also, these instructions are technically "assembled" by an "assembler". Basically the Risc-V assembler takes the instructions (or mnemonics) and encodes them into binary words which are "fed" to the cpu. The instructions that an assembler understands make up an "assembly language". Compiling is a term used for higher level languages like C/C++.

– dopamane
Nov 15 '18 at 8:04

|
show 1 more comment

Take the instructions one at a time. First the load-upper-immediate, take the immediate (0x1234) and "load" it into the upper half of the S0 register and zero out the lower half:

lui S0, 0x1234 

S0 = 0x12340000

Next the or-immediate, we OR the value in S0 with the value 0x5678:

ori S1, S0, 0x5678

 0x12340000
OR 0x00005678
 ----------
 0x12345678 = S1

Finally the add, we are adding the value in S1 to itself or, equivalently, multiplying the value in S1 by 2:

add S2, S1, S1

 0x12345678
+ 0x12345678
 ----------
 0x2468ACF0 = S2

edited Nov 15 '18 at 2:56

answered Nov 15 '18 at 2:50

dopamane

85611024

Take the instructions one at a time. First the load-upper-immediate, take the immediate (0x1234) and "load" it into the upper half of the S0 register and zero out the lower half:

lui S0, 0x1234 

S0 = 0x12340000

Next the or-immediate, we OR the value in S0 with the value 0x5678:

ori S1, S0, 0x5678

 0x12340000
OR 0x00005678
 ----------
 0x12345678 = S1

Finally the add, we are adding the value in S1 to itself or, equivalently, multiplying the value in S1 by 2:

add S2, S1, S1

 0x12345678
+ 0x12345678
 ----------
 0x2468ACF0 = S2

edited Nov 15 '18 at 2:56

answered Nov 15 '18 at 2:50

dopamane

85611024

edited Nov 15 '18 at 2:56

answered Nov 15 '18 at 2:50

dopamane

85611024

answered Nov 15 '18 at 2:50

dopamane

85611024

answered Nov 15 '18 at 2:50

dopamane

85611024

thank you so much sir , finally i understand how to compile thanks to you. very useful

– Razi Awad
Nov 15 '18 at 7:39

can i ask you one tiny question ? i have read that the lui instruction moves the 16 bits to the higher and the other 16 bits to the lower . how did you determine which side is higher and which is lower . also the 0x1234 does it contain 4 bits ? (1234) i have problem understanding this

– Razi Awad
Nov 15 '18 at 7:42

becuase we should have written it like that : 0x1234 0000 0000 0000 0000 0000 0000 0000 which is 32 bits half of it higher whiich is far left and half of it lower far right , Is it like that ?

– Razi Awad
Nov 15 '18 at 7:45

0x1234 is base 16 (or hexadecimal), so each "digit" is four bits. 0x1234 (base 16) = 0001_0010_0011_0100 (base 2). A register has 32-bits or 8 hex digits. The spec fig2.1 shows the most-significant-bit (XLEN-1) to the left-side. So upper is left, lower is right.

– dopamane
Nov 15 '18 at 7:52

Also, these instructions are technically "assembled" by an "assembler". Basically the Risc-V assembler takes the instructions (or mnemonics) and encodes them into binary words which are "fed" to the cpu. The instructions that an assembler understands make up an "assembly language". Compiling is a term used for higher level languages like C/C++.

– dopamane
Nov 15 '18 at 8:04

|
show 1 more comment

thank you so much sir , finally i understand how to compile thanks to you. very useful

– Razi Awad
Nov 15 '18 at 7:39

can i ask you one tiny question ? i have read that the lui instruction moves the 16 bits to the higher and the other 16 bits to the lower . how did you determine which side is higher and which is lower . also the 0x1234 does it contain 4 bits ? (1234) i have problem understanding this

– Razi Awad
Nov 15 '18 at 7:42

becuase we should have written it like that : 0x1234 0000 0000 0000 0000 0000 0000 0000 which is 32 bits half of it higher whiich is far left and half of it lower far right , Is it like that ?

– Razi Awad
Nov 15 '18 at 7:45

0x1234 is base 16 (or hexadecimal), so each "digit" is four bits. 0x1234 (base 16) = 0001_0010_0011_0100 (base 2). A register has 32-bits or 8 hex digits. The spec fig2.1 shows the most-significant-bit (XLEN-1) to the left-side. So upper is left, lower is right.

– dopamane
Nov 15 '18 at 7:52

Also, these instructions are technically "assembled" by an "assembler". Basically the Risc-V assembler takes the instructions (or mnemonics) and encodes them into binary words which are "fed" to the cpu. The instructions that an assembler understands make up an "assembly language". Compiling is a term used for higher level languages like C/C++.

– dopamane
Nov 15 '18 at 8:04

thank you so much sir , finally i understand how to compile thanks to you. very useful

– Razi Awad
Nov 15 '18 at 7:39

can i ask you one tiny question ? i have read that the lui instruction moves the 16 bits to the higher and the other 16 bits to the lower . how did you determine which side is higher and which is lower . also the 0x1234 does it contain 4 bits ? (1234) i have problem understanding this

– Razi Awad
Nov 15 '18 at 7:42

becuase we should have written it like that : 0x1234 0000 0000 0000 0000 0000 0000 0000 which is 32 bits half of it higher whiich is far left and half of it lower far right , Is it like that ?

– Razi Awad
Nov 15 '18 at 7:45

0x1234 is base 16 (or hexadecimal), so each "digit" is four bits. 0x1234 (base 16) = 0001_0010_0011_0100 (base 2). A register has 32-bits or 8 hex digits. The spec fig2.1 shows the most-significant-bit (XLEN-1) to the left-side. So upper is left, lower is right.

– dopamane
Nov 15 '18 at 7:52

Also, these instructions are technically "assembled" by an "assembler". Basically the Risc-V assembler takes the instructions (or mnemonics) and encodes them into binary words which are "fed" to the cpu. The instructions that an assembler understands make up an "assembly language". Compiling is a term used for higher level languages like C/C++.

– dopamane
Nov 15 '18 at 8:04

|
show 1 more comment

"LUI places the U-immediate value in the top 20 bits of the destination register rd, filling in the lowest 12 bits with zeros." -- riscv-spec-v2.2

so,
lui s0, 0x1234

s0 should be 0x01234000

edited Nov 23 '18 at 2:54

answered Nov 23 '18 at 2:48

jimmy li

add a comment |

"LUI places the U-immediate value in the top 20 bits of the destination register rd, filling in the lowest 12 bits with zeros." -- riscv-spec-v2.2

so,
lui s0, 0x1234

s0 should be 0x01234000

edited Nov 23 '18 at 2:54

answered Nov 23 '18 at 2:48

jimmy li

add a comment |

"LUI places the U-immediate value in the top 20 bits of the destination register rd, filling in the lowest 12 bits with zeros." -- riscv-spec-v2.2

so,
lui s0, 0x1234

s0 should be 0x01234000

edited Nov 23 '18 at 2:54

answered Nov 23 '18 at 2:48

jimmy li

"LUI places the U-immediate value in the top 20 bits of the destination register rd, filling in the lowest 12 bits with zeros." -- riscv-spec-v2.2

so,
lui s0, 0x1234

s0 should be 0x01234000

edited Nov 23 '18 at 2:54

answered Nov 23 '18 at 2:48

jimmy li

edited Nov 23 '18 at 2:54

answered Nov 23 '18 at 2:48

jimmy li

answered Nov 23 '18 at 2:48

jimmy li

answered Nov 23 '18 at 2:48

jimmy li

add a comment |

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