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x86汇编 - 使用加法查找权力(x86 Assembly - finding powers using addition)

 该程序将接受来自用户的两个数字并显示这两个数字的总和，乘积和功率（a ^ b）。 这是捕获，但是...... 
 

 程序必须：使用AddNumbers函数在MultiplyNumbers函数中使用AddNumbers函数在CalculatePower函数中使用MultiplyNumbers函数 
 

 我无法通过实现乘法加法功能找出如何找到两个数的乘积。 我无法理解它背后的逻辑。 我显然做错了什么，但我不确定是什么。 （请忽略代码的低效率，我只想知道在找到两者的产品时我做错了什么） 
 

 我的代码到目前为止.. 
 

INCLUDE Irvine32.inc
.data
str1 BYTE "Enter a positive integer: ",0
str2 BYTE "The sum is: ",0
str3 BYTE "The product is: ",0
str4 BYTE "The power result is: ",0

.code
main PROC

call GetInteger
call Crlf
mov eax, ebx
call AddNumbers

mov edx, OFFSET str2
call WriteString
call WriteInt
call Crlf
mov eax, 0
mov ecx, edi
call MultiplyNumber
mov eax, edx
mov edx, OFFSET str3
call WriteString
call WriteInt
call Crlf
call CalculatePower
mov eax, esi
mov edx, OFFSET str4
call WriteString
call WriteInt
call Crlf

exit
main ENDP

GetInteger PROC 
mov edx, OFFSET str1
call WriteString
call ReadInt
mov ebx, eax
call WriteString
call ReadInt
mov edi, eax
ret
GetInteger ENDP

CalculatePower PROC USES edi ebx 

mov ecx, edi
mov esi, 0
L2:

    call MultiplyNumber
    add esi, edx
    loop L2
    ret
CalculatePower ENDP

MultiplyNumber PROC USES ebx edi ecx

mov edx, 0
L1:
    mov edi, ebx
    mov eax, 0
    call AddNumbers
    add edx, eax
    loop L1
    ret
MultiplyNumber ENDP

AddNumbers PROC 

    add eax, edi

ret
AddNumbers ENDP

END main


The program will accept two numbers from the user and display the sum, product, and power (a^b) of those two numbers. Here is the catch, however...
 

The program MUST: Use an AddNumbers function Use that AddNumbers function in a MultiplyNumbers function Use that MultiplyNumbers function in a CalculatePower function
 

I cannot figure out how to find the Product of the two numbers by implementing the multiplication through addition function. I cannot grasp the logic behind it. I'm obviously doing something wrong, but I'm not sure what. (Please ignore the inefficiency of the code, I just want to know what I'm doing wrong as far as finding the product of the two)
 

My code so far..
 

INCLUDE Irvine32.inc
.data
str1 BYTE "Enter a positive integer: ",0
str2 BYTE "The sum is: ",0
str3 BYTE "The product is: ",0
str4 BYTE "The power result is: ",0

.code
main PROC

call GetInteger
call Crlf
mov eax, ebx
call AddNumbers

mov edx, OFFSET str2
call WriteString
call WriteInt
call Crlf
mov eax, 0
mov ecx, edi
call MultiplyNumber
mov eax, edx
mov edx, OFFSET str3
call WriteString
call WriteInt
call Crlf
call CalculatePower
mov eax, esi
mov edx, OFFSET str4
call WriteString
call WriteInt
call Crlf

exit
main ENDP

GetInteger PROC 
mov edx, OFFSET str1
call WriteString
call ReadInt
mov ebx, eax
call WriteString
call ReadInt
mov edi, eax
ret
GetInteger ENDP

CalculatePower PROC USES edi ebx 

mov ecx, edi
mov esi, 0
L2:

    call MultiplyNumber
    add esi, edx
    loop L2
    ret
CalculatePower ENDP

MultiplyNumber PROC USES ebx edi ecx

mov edx, 0
L1:
    mov edi, ebx
    mov eax, 0
    call AddNumbers
    add edx, eax
    loop L1
    ret
MultiplyNumber ENDP

AddNumbers PROC 

    add eax, edi

ret
AddNumbers ENDP

END main


原文：https://stackoverflow.com/questions/25840822
更新时间：2023-04-15 17:04

最满意答案

 不幸的是，不，您必须为每个类别再次指定红色，绿色和蓝色值： 
 

a { display: block; position: relative; }

.brown { position: absolute; bottom: 0; background-color: rgba(118, 76, 41, 0.8); }
a:hover .brown { background-color: rgba(118, 76, 41, 1); }

.green { position: absolute; bottom: 0; background-color: rgba(51, 91, 11, 0.8); }
a:hover .green { background-color: rgba(51, 91, 11, 1); }

 

 您只能使用inherit关键字作为属性的值，即使在此处使用inherit也不适用。 

This is now possible with custom properties:
 

.brown { --rgb: 118, 76, 41; }
.green { --rgb: 51, 91, 11; }

a { display: block; position: relative; }
div { position: absolute; bottom: 0; background-color: rgba(var(--rgb), 0.8); }
a:hover div { background-color: rgba(var(--rgb), 1); }

 

To understand how this works, see How do I apply opacity to a CSS color variable?
 

If custom properties are not an option, see the original answer below.
 

 

Unfortunately, no, you'll have to specify the red, green and blue values again for each individual class:
 

a { display: block; position: relative; }

.brown { position: absolute; bottom: 0; background-color: rgba(118, 76, 41, 0.8); }
a:hover .brown { background-color: rgba(118, 76, 41, 1); }

.green { position: absolute; bottom: 0; background-color: rgba(51, 91, 11, 0.8); }
a:hover .green { background-color: rgba(51, 91, 11, 1); }

 

You can only use the inherit keyword alone as a value for the property, and even then the use of inherit isn't appropriate here.

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