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使用加法器电路添加两个4位二进制数

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我需要编写一个程序,使用加法器电路添加两个4位二进制数字,为我们提供答案 . 我需要使用int()函数将其从二进制转换为十进制(两个值都需要显示) . 我正在努力创建一个创建正确的二进制和十进制输出的代码 . 我们输入两组四位数字 . 数字应该通过电路,结果应该加在一起 . 现在大门都很好 . 但是,我无法正确地将数字加在一起 . 通过电路的二进制加法是不正确的 . 转换器和输入都是正确的 .

用户需要为x(x1,x2,x3,x4)和y(y1,y2,y3,y4)和ci = 0输入4位二进制数 . 例如:x = 1111 y = 0000

This is the adder circuit along with the diagram that shows how the circuit will look after adding the two binary numbers(I can't embed images yet)

这是我目前的代码:

import string

print('Simple Circuits for Final Project')

x = input('x| enter 4 digits: ')
y = input('y| enter 4 digits: ')

list1 = list(x)
print(list1)

list2 = list(y)
print(list2)



#define all the gates
def xorgate(x,y):
    if x == y:
        return '0'
    else:
        return '1'

def andgate (x,y):
    if x == '1' and y == '1':
        return '1'
    else:
        return '0'

def orgate (x,y):
    if x == '1' or y == '1':
        return '1'
    else:
        return '0'

#define the entire circuit and input the list of inputs for the circuit.
#include the outputs based on the gates defined above.
def circuit(x,y,ci):
    a = 3 #starting value for the while loop, get approp position for our list
    adder = ['0','0','0','0','0']#adder list only has strings of zero. list to hold binary numbers from the output of the adding circuit
    b = 4
    while a >= 0:



        xorout = xorgate(x[a],y[a])
        print("xor:",(xorout))
        s = xorgate(ci,xorout)
        print("Ci:",ci)
        andout = andgate(xorout, ci)
        print("and1:",andout)
        and2out = andgate(x[a],y[a])
        print("and2:",and2out)
        co = orgate(andout,and2out)

        print('s:',s)
        print('co:',co)
        print('-----------------')

        ci = co
        adder[b] = str(s)
        a-=1
        b-=1


    adder[4]=ci
    #print(final)
    print('The Final Binary Output is:', adder[::-1])

    #OUR CONVERTER IS RIGHT, BUT WE CAN'T GET THE BINARY ADDITION ACCURATE
    outputfinal = ((int(adder[4]) * 16) + (int(adder[3]) * 8)+(int(adder[2]) * 4)+(int(adder[1]) * 2)+(int(adder[0]) * 1))
    print('The Final Decimal Output is:', outputfinal)
hold = '0'
circuit(list1,list2,hold)# 3 value for circuit

这是我们认为错误的部分:

ci = co
    adder[b] = str(s)
    a-=1
    b-=1


adder[4]=ci
#print(final)
print('The Final Binary Output is:', adder[::-1])

这是我当前的输出,这是错误的:

x| enter 4 digits: 1111
y| enter 4 digits: 0000
['1', '1', '1', '1']
['0', '0', '0', '0']
xor: 1
Ci: 0
and1: 0
and2: 0
s: 1
co: 0
-----------------
xor: 1
Ci: 0
and1: 0
and2: 0
s: 1
co: 0
-----------------
xor: 1
Ci: 0
and1: 0
and2: 0
s: 1
co: 0
-----------------
xor: 1
Ci: 0
and1: 0
and2: 0
s: 1
co: 0
-----------------
The Final Binary Output is: ['0', '1', '1', '1', '0']
The Final Decimal Output is: 14

1 回答

  • 2

    您混淆了加法器列表的顺序 .

    应该是第一个地址是0,而不是4:

    adder[0]=ci
    #print(final)
    

    不要反转列表

    print('The Final Binary Output is:', adder)
    

    由于你的转换器期望它的顺序相反,我在这里反转它而不是重写你的转换器:

    adder = adder[::-1]
    #OUR CONVERTER IS RIGHT, BUT WE CAN'T GET THE BINARY ADDITION ACCURATE
    outputfinal = ((int(adder[4]) * 16) + (int(adder[3]) * 8)+(int(adder[2]) * 4)+(int(adder[1]) * 2)+(int(adder[0]) * 1))
    print('The Final Decimal Output is:', outputfinal)
    

    您可以在在线代表中试用整个程序(因为您得到了不同的结果):https://repl.it/repls/CompetitiveSerpentineAccess

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