Help,design a 4 bit adder.The program should convert 2 decimal to binary then add the 2 giving a sum and carry as the final result.

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C#

#include<iostream>
using namespace std;
int main()
{

bool ar[4],arr[4] ;
int i=0,h=0,j,k;
int n,m,carry=0,result[5];
cout<<"Enter the first positive number:"<<"\n";
cin>>n;
while(n>0)
{
    ar[i]=n%2;
    i++;
    n=n/2;

}

cout<<"Binary number is:";
for (j=i-1;j>=0;j--)
{
    cout<<ar[j];
    }
cout<<"\n"<<"Enter the 2nd positive number:"<<"\n";
cin>>m;
while(m>0)
{
    arr[h]=m%2;
    h++;
    m=m/2;

}

cout<<"Binary number is:";
for (k=h-1;k>=0;k--)
{
    cout<<arr[k];
    }

    cout<<"\n";

Posted 4-Oct-15 9:58am

Mpaphi Map'z Dambe

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Comments

CHill60 4-Oct-15 16:35pm

What is your question?

Mpaphi Map'z Dambe 4-Oct-15 16:51pm

the question was to design a 4 bit adder .
It should convert two decimal to 4 bit binary then add the two binary numbers.for example 9 and 10.9=1001 and 10=1010 adding the two will result in 10011 carry=1

CHill60 4-Oct-15 17:08pm

No - that's your homework assignment.
You've posted some code - what's wrong with it.

Mpaphi Map'z Dambe 4-Oct-15 17:49pm

Im stuck
i was able to make a code that can convert a decimal to binary(storing each bit in an array)but now im failing to produce a code that will add the two array...thats what i need help with

2 solutions

Solution 1

The code you provided is already buggy: ar and arr are not initialised to zero.
One way to correct it is:

C#

for(i=0; i < 4, i++)
{
    ar[i]=n%2;
    n=n/2;
}

This code will fill ar and zero it at the same time.

What have you done about the bits adder ?
By the way we don't do your homework for you, we just help if you have a problem in the work you have done.

Posted 4-Oct-15 11:12am

Patrice T

Updated 4-Oct-15 11:18am

v2

Comments

Mpaphi Map'z Dambe 4-Oct-15 17:56pm

that part is sorted i used a while function while(n>0)
{
ar[i]=n%2;
i++;
n=n/2;
you can run the code,its working(so far it can convert decimal to binary),im struggling to add the two array ar[] and arr[] .
i need to code such that
ar[3] +arr[3]=result[3]
ar[2]+arr[2]=result[2]
ar[1]+arr[1]=result[1]
ar[0]+arr[0]=result[0]

which will give my final result[5]

Patrice T 4-Oct-15 18:22pm

I fear you can't use an addition to make an adder.

Matt T Heffron 5-Oct-15 12:04pm

Hint: XOR is a half-adder and AND is for the carry.

Mpaphi Map'z Dambe 4-Oct-15 18:43pm

Thats what im struggling with, an operation or function to use to add the 2 arrays

Patrice T 4-Oct-15 18:53pm

What is your code to add 2 bits ?

Mpaphi Map'z Dambe 5-Oct-15 7:10am

ppolymorphe thats what i nid help on caz the code i did was giving me somthn else

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Andreas Gieriet · Accepted Answer · 2015-10-04T13:40:00

Solution 2

From your description it is not obvious how much you need to implement in terms of hardware-like full adder? If it's just about producing the result, you may very much simplify the code, e.g. (not tried with the compiler - from brain to paper ;-)):

C++

#include <cctype>
#include <iostream>
using namespace std;
int main()
{
    unsigned int a;
    unsigned int b;
    cin >> a >> b;
    if (a < 0x10 && b < 0x10)
    {
        unsigned int r = a + b;
        unsigned int c = r & 0x10 ? 1 : 0;
        r &= 0x0F;
        cout << "carry = " << c << ", 4-bit result = " << r << endl;
    }
}

Since it was neither clear if you need to dump the result as binary or as decimal, I decided for the least effort: decimal ;-).
Regards
Andi

Posted 4-Oct-15 13:40pm

Andreas Gieriet

Updated 4-Oct-15 13:42pm

v3

Comments

Patrice T 4-Oct-15 20:05pm

I think that building a 4 bits adder should exclude using the addition.

Andreas Gieriet 5-Oct-15 0:21am

Well, that's reading between the lines ;-). My point is, that if the "requirement" is not clear enough, the solution will be just "anything".
Regards
Andi
PS: But you are right, the intent was maybe to mimic hardware implementation.

Patrice T 5-Oct-15 1:49am

:)

Mpaphi Map'z Dambe 5-Oct-15 7:12am

Andreas the code has to add 2 4bit numbers.first it has to convert 2 decimal to binary then add them.

Andreas Gieriet 5-Oct-15 15:04pm

You don't have to repeat your title here again. It is obviously ambiguous, or more precisely: leaves open crucial "details". May I use the plus operator? in which form is the output expected (you mention that you enter decimal numbers to be converted to binary - so, the result is decimal again - or binary?). What do you need in the end? Mimic hardware? E.g. have a 4-bit adder build out of four full adders connected by carries, and each full adder composed by two half adders plus an or gate? We cannot know what the goal of this exercise is, what your background is, etc.
A possible implementation, given your C++ tag is to create components (classes) in C++ which represent the various pieces of hardware and communicate over variables (which may play the role of wires). What is the whole context? What are you expected to learn from this exercise?
If it's about mimicking hardware, I suggest to at least make functions for the individual functional parts.
Regards
Andi

Mpaphi Map'z Dambe 5-Oct-15 7:16am

and i ran the code its not given the correct result

Andreas Gieriet 5-Oct-15 14:56pm

What is the correct result for e.g. 9 and 11? for me: carry = 1, result = 4. I.e. a *4* bit adder has a *4* bit result *plus* a carry flag that indicates - a carry.
Regards
Andi

Mpaphi Map'z Dambe 5-Oct-15 15:10pm

9+11 should give 20 as the ansa...and 1 as yo carry

Andreas Gieriet 5-Oct-15 15:14pm

A 4 bit adder in hardware has always a *4* bit result. The carry is the indicator that a fifth bit was needed to represent the full result.
Or am I missing something?
Take any CPU (which has an ALU) which has only N-bit registers (e.g. 32 bits). Unsigned adding two 32 bit registers results in a 32 bit value *plus* an indicator if there was a carry (i.e. the result cannot be fully represented in 32 bits).
Regards
Andi

Mpaphi Map'z Dambe 5-Oct-15 15:14pm

So how can i make it to add two 4 bit numbers that would give a result of a 5 bit?

Andreas Gieriet 5-Oct-15 15:16pm

What does your text book say about this?

Mpaphi Map'z Dambe 5-Oct-15 15:21pm

thanks alot i think now i understand,the final result is a sum and a carry .

Mpaphi Map'z Dambe 5-Oct-15 15:23pm

i was confused by your cout"4 bit result" instead it should read"sum"

Andreas Gieriet 5-Oct-15 15:25pm

Well, I deliberately wrote "4 bit" since the sum *is* 4 bits and the carry is the potential 5th bit.

Mpaphi Map'z Dambe 5-Oct-15 15:30pm

I see
thanks though