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julian salas
Asked: 2020-02-18 08:38:19 +0800 CST 2020-02-18 08:38:19 +0800 CST

Add two variables in MASM Assembler

  • 772

I have a question about the sum in registers for MASM Assembler

 TITLE Suma variables
 INCLUDE Irvine32.inc
 .data
 a dword 10000h
 b dword 40000h
 valorFinal dword ?
.code
 main PROC
       mov eax,a ; empieza con 10000h
       add eax,b ; suma 40000h
       mov valorFinal,eax ; 
      call DumpRegs
      exit
 main ENDP
 END main

My doubt is when I use addwith b, am I adding only the value of the variable, or am I adding the value and the address in memory, because I understand that to obtain the specific value it must be enclosed in [].

ensamblador

2 Answers

  1. Best Answer

    In MASM (and TASM in MASM-compatible mode), when you type:

    a dword 10000h
b dword 40000h

    aand bare labels that represent the address of the storage assigned to the double words 10000hand 40000hrespectively.

    When a label is used as an operand, MASM knows that the label represents a memory address, and decides that the parameter is a memory reference. To force the memory address to be used as an immediate value, it can be used OFFSETpreceding the tag. Summarizing:

    mov eax, a          ; mueve 10000h a eax
mov eax, OFFSET a   ; mueve la dirección donde está 10000h a eax
mov eax, [a]        ; en MASM equivale a MOV eax, a

    Instead, when you write:

    foo EQU 42
bar EQU 66

    fooand barare symbolic constants that represent 42 and 66, with no space reserved for 42 and 66. When using a symbolic constant as a parameter, MASM treats the parameter as an immediate value.

    mov eax, foo        ; mueve 42 a eax

    The fact that the meaning of a statement is different depending on whether its parameters are labels or constants can be confusing. Other assemblers, such as NASM (also Yasm, which uses NASM syntax), TASM in IDEAL mode, or fasm, require that brackets be used to treat a parameter as a memory reference, and if there are no brackets they always treat the parameter as an immediate value.

    • 9

  2. With the instruction a dword 10000hyou are defining a 4-byte memory zone that contains the value 10000h, in this case ait refers to the address of that memory zone (which will be the one that the compiler decides) and with mov eax,athe accumulator you are loading the address memory, not the contents of memory.

    It's not clear from your question what you want to achieve, but it will probably be one of two things:

    1. Load the accumulator with the value stored in the memory area defined by a(ie, have the processor read 4 bytes from aand store the result in eax). In that case you should use mov eax,[a].

    2. Define aas a constant, which will be directly replaced by the corresponding value at compile time. In that case you should define the constant as a equ 10000h(don't use dword). The compiled code will then be equivalent to mov eax,10000h.

    And the same applies to instruction add.

    • 7