“operands do not match” assembler syntax error

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I wrote this code, in assembly, to calculate (4*7)+5, but I get these errors:

(18) wrong parameters: MOV ax,m1

(18) operands do not match: 16 bit register and 8 bit address.

 .MODEL SMALL
.stack 100h
.data
m1 DB 4
m2 DB 7
m3 DB 5
m4 DB ?
.code
main PROC
 mov ax,@data
 mov ds,ax
 mov ax,m1
 mul ax,m2
 add ax,m3
 mov m4,ax 

 BEGIN ENDP
 END main

edited Nov 11 at 18:24

Peter Cordes

117k16178305

asked Nov 11 at 17:19

Faisal

113

2

@movcmpret: no it won't. This is x86 with MASM-like syntax. The error is that symbol names magically get a size associated with them when declared that way, and mov ax, m1 is doing a word load from a byte "variable", and the assembler stops you from making that mistake. One solution is to use dw and make everything a word, or to use byte registers (al)
– Peter Cordes
Nov 11 at 17:26

I am a beginner in the assembly language, can you help me?
– Faisal
Nov 11 at 17:26

I just edited my last comment.
– Peter Cordes
Nov 11 at 17:27

I will trying now
– Faisal
Nov 11 at 17:34

I do it. But If i running the program I didn't now where i can see the result.
– Faisal
Nov 11 at 17:51

|
show 8 more comments

up vote
2
down vote

favorite

I wrote this code, in assembly, to calculate (4*7)+5, but I get these errors:

(18) wrong parameters: MOV ax,m1

(18) operands do not match: 16 bit register and 8 bit address.

 .MODEL SMALL
.stack 100h
.data
m1 DB 4
m2 DB 7
m3 DB 5
m4 DB ?
.code
main PROC
 mov ax,@data
 mov ds,ax
 mov ax,m1
 mul ax,m2
 add ax,m3
 mov m4,ax 

 BEGIN ENDP
 END main

edited Nov 11 at 18:24

Peter Cordes

117k16178305

asked Nov 11 at 17:19

Faisal

113

2

@movcmpret: no it won't. This is x86 with MASM-like syntax. The error is that symbol names magically get a size associated with them when declared that way, and mov ax, m1 is doing a word load from a byte "variable", and the assembler stops you from making that mistake. One solution is to use dw and make everything a word, or to use byte registers (al)
– Peter Cordes
Nov 11 at 17:26

I am a beginner in the assembly language, can you help me?
– Faisal
Nov 11 at 17:26

I just edited my last comment.
– Peter Cordes
Nov 11 at 17:27

I will trying now
– Faisal
Nov 11 at 17:34

I do it. But If i running the program I didn't now where i can see the result.
– Faisal
Nov 11 at 17:51

|
show 8 more comments

up vote
2
down vote

favorite

I wrote this code, in assembly, to calculate (4*7)+5, but I get these errors:

(18) wrong parameters: MOV ax,m1

(18) operands do not match: 16 bit register and 8 bit address.

 .MODEL SMALL
.stack 100h
.data
m1 DB 4
m2 DB 7
m3 DB 5
m4 DB ?
.code
main PROC
 mov ax,@data
 mov ds,ax
 mov ax,m1
 mul ax,m2
 add ax,m3
 mov m4,ax 

 BEGIN ENDP
 END main

edited Nov 11 at 18:24

Peter Cordes

117k16178305

asked Nov 11 at 17:19

Faisal

113

I wrote this code, in assembly, to calculate (4*7)+5, but I get these errors:

(18) wrong parameters: MOV ax,m1

(18) operands do not match: 16 bit register and 8 bit address.

 .MODEL SMALL
.stack 100h
.data
m1 DB 4
m2 DB 7
m3 DB 5
m4 DB ?
.code
main PROC
 mov ax,@data
 mov ds,ax
 mov ax,m1
 mul ax,m2
 add ax,m3
 mov m4,ax 

 BEGIN ENDP
 END main

assembly x86-16

edited Nov 11 at 18:24

Peter Cordes

117k16178305

asked Nov 11 at 17:19

Faisal

113

edited Nov 11 at 18:24

Peter Cordes

117k16178305

asked Nov 11 at 17:19

Faisal

113

edited Nov 11 at 18:24

Peter Cordes

117k16178305

edited Nov 11 at 18:24

Peter Cordes

117k16178305

edited Nov 11 at 18:24

Peter Cordes

117k16178305

asked Nov 11 at 17:19

Faisal

113

asked Nov 11 at 17:19

Faisal

113

asked Nov 11 at 17:19

Faisal

113

2

@movcmpret: no it won't. This is x86 with MASM-like syntax. The error is that symbol names magically get a size associated with them when declared that way, and mov ax, m1 is doing a word load from a byte "variable", and the assembler stops you from making that mistake. One solution is to use dw and make everything a word, or to use byte registers (al)
– Peter Cordes
Nov 11 at 17:26

I am a beginner in the assembly language, can you help me?
– Faisal
Nov 11 at 17:26

I just edited my last comment.
– Peter Cordes
Nov 11 at 17:27

I will trying now
– Faisal
Nov 11 at 17:34

I do it. But If i running the program I didn't now where i can see the result.
– Faisal
Nov 11 at 17:51

|
show 8 more comments

2

@movcmpret: no it won't. This is x86 with MASM-like syntax. The error is that symbol names magically get a size associated with them when declared that way, and mov ax, m1 is doing a word load from a byte "variable", and the assembler stops you from making that mistake. One solution is to use dw and make everything a word, or to use byte registers (al)
– Peter Cordes
Nov 11 at 17:26

I am a beginner in the assembly language, can you help me?
– Faisal
Nov 11 at 17:26

I just edited my last comment.
– Peter Cordes
Nov 11 at 17:27

I will trying now
– Faisal
Nov 11 at 17:34

I do it. But If i running the program I didn't now where i can see the result.
– Faisal
Nov 11 at 17:51

@movcmpret: no it won't. This is x86 with MASM-like syntax. The error is that symbol names magically get a size associated with them when declared that way, and mov ax, m1 is doing a word load from a byte "variable", and the assembler stops you from making that mistake. One solution is to use dw and make everything a word, or to use byte registers (al)
– Peter Cordes
Nov 11 at 17:26

I am a beginner in the assembly language, can you help me?
– Faisal
Nov 11 at 17:26

I just edited my last comment.
– Peter Cordes
Nov 11 at 17:27

I will trying now
– Faisal
Nov 11 at 17:34

I do it. But If i running the program I didn't now where i can see the result.
– Faisal
Nov 11 at 17:51

|
show 8 more comments

1 Answer
1

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votes

up vote
2
down vote

Because all of the numbers involved 4,7,5 are very small, you get by with data definitions using db.
But this also means that the calculations should be byte-sized.

Write this instead:

mov al, m1
mul m2 ;Leaves m1 * m2 in AX (4*7=28 so AL=28 and AH=0)
add al, m3
mov m4, al

If you use a byte-sized operand (m2) with the mul instruction then AL is multiplied with it and the double-length product is stored in AX.

answered Nov 11 at 18:35

Sep Roland

11.6k21845

Thank you.......
– Faisal
Nov 11 at 18:40

add a comment |

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1 Answer
1

active

oldest

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1 Answer
1

active

oldest

votes

up vote
2
down vote

Because all of the numbers involved 4,7,5 are very small, you get by with data definitions using db.
But this also means that the calculations should be byte-sized.

Write this instead:

mov al, m1
mul m2 ;Leaves m1 * m2 in AX (4*7=28 so AL=28 and AH=0)
add al, m3
mov m4, al

If you use a byte-sized operand (m2) with the mul instruction then AL is multiplied with it and the double-length product is stored in AX.

answered Nov 11 at 18:35

Sep Roland

11.6k21845

Thank you.......
– Faisal
Nov 11 at 18:40

add a comment |

up vote
2
down vote

Because all of the numbers involved 4,7,5 are very small, you get by with data definitions using db.
But this also means that the calculations should be byte-sized.

Write this instead:

mov al, m1
mul m2 ;Leaves m1 * m2 in AX (4*7=28 so AL=28 and AH=0)
add al, m3
mov m4, al

If you use a byte-sized operand (m2) with the mul instruction then AL is multiplied with it and the double-length product is stored in AX.

answered Nov 11 at 18:35

Sep Roland

11.6k21845

Thank you.......
– Faisal
Nov 11 at 18:40

add a comment |

up vote
2
down vote

Because all of the numbers involved 4,7,5 are very small, you get by with data definitions using db.
But this also means that the calculations should be byte-sized.

Write this instead:

mov al, m1
mul m2 ;Leaves m1 * m2 in AX (4*7=28 so AL=28 and AH=0)
add al, m3
mov m4, al

If you use a byte-sized operand (m2) with the mul instruction then AL is multiplied with it and the double-length product is stored in AX.

answered Nov 11 at 18:35

Sep Roland

11.6k21845

Because all of the numbers involved 4,7,5 are very small, you get by with data definitions using db.
But this also means that the calculations should be byte-sized.

Write this instead:

mov al, m1
mul m2 ;Leaves m1 * m2 in AX (4*7=28 so AL=28 and AH=0)
add al, m3
mov m4, al

If you use a byte-sized operand (m2) with the mul instruction then AL is multiplied with it and the double-length product is stored in AX.

answered Nov 11 at 18:35

Sep Roland

11.6k21845

answered Nov 11 at 18:35

Sep Roland

11.6k21845

answered Nov 11 at 18:35

Sep Roland

11.6k21845

answered Nov 11 at 18:35

Sep Roland

11.6k21845

Thank you.......
– Faisal
Nov 11 at 18:40

add a comment |

Thank you.......
– Faisal
Nov 11 at 18:40

Thank you.......
– Faisal
Nov 11 at 18:40

add a comment |

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