how to use raise to the power in c++

Question

1.00/5 (2 votes)

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i am trying to convert binary to decimal by this code the problem is that how to use 2 raise to the power -

C++

#include<iostream>
using namespace std;
#include<conio.h>

int main()
{
	int a,b,temp=0,n;
	cout<<"Enter the range of natural no's : ";
	cin>>b;
	n==b;
	for(a=0;a<=n;a++)
	{
		temp=temp+n%10;
		temp=temp*2^a;
		n=n/10;
	}
	cout<<"\nThe sum of given natural no's is : "<<temp;
	getch();
	return 0;
}

Posted 26-Jun-15 16:14pm

nihal saxena

Updated 28-Jun-15 16:24pm

v4

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Comments

[no name] 26-Jun-15 22:23pm

http://www.cplusplus.com/reference/cmath/pow/

Sergey Alexandrovich Kryukov 26-Jun-15 23:37pm

No, no. Pay attention that the operands are integer. Please see my answer.
—SA

[no name] 27-Jun-15 6:04am

No, no yourself. Pay attention to the question.

Sergey Alexandrovich Kryukov 27-Jun-15 8:04am

Why? What part of the question? Do you mean "decimal to binary"? Still no floating-point arithmetic...
But I need to update the answer.
—SA

CPallini 27-Jun-15 4:47am

What's exactly your requirement? It doesn't look a decimal to binary conversion.

4 solutions

Solution 2

In addition to Solution 1, watch the following problem:

C++

n==b;

Compares n and b and throws the result away! You probably meant to write:

C++

n = b;

Posted 26-Jun-15 19:28pm

nv3

Comments

[no name] 27-Jun-15 8:35am

a 5 for the eagle eye :)

nv3 27-Jun-15 14:32pm

Thanks!

Solution 4

If the question is, how to print the memory image of some integer value, you might use bitset. E.g.

C++

#include <bitset>
...
int x = ...
...
std::cout << "The binary value of x = " << x
          << " is " << std::bitset<sizeof(int)*8>(x)
          << std::endl;

See also http://www.cplusplus.com/reference/bitset/bitset/bitset/[^].
Cheers
Andi

Posted 28-Jun-15 1:23am

Andreas Gieriet

Solution 3

Sure, 1st you want to add this to the top of your code:

C++

#include <cmath>

As for your power, there's a function for that:

C++

pow(5, 3);

Hope I helped you out some :)

Posted 27-Jun-15 2:41am

Jennie Smith

Updated 28-Jun-15 1:04am

v2

Comments

Afzaal Ahmad Zeeshan 27-Jun-15 13:38pm

Perhaps you didn't read the question. He said he wants to convert decimal to binary.

Andreas Gieriet 28-Jun-15 7:05am

I fixed the preformatted code blocks.
But as the comment above mentions, this is not a useful answer to the question.
Cheers
Andi

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Sergey Alexandrovich Kryukov · Accepted Answer · 2015-06-26T17:36:00

Solution 1

Let's note that the operands are integers. The fastest way to find the integer power a of 2 is

C++

int power = 1 << a;

The method suggested by Wes Aday in his comment to the question is only good for floating point calculation; for integers it would involve double type cast — not good.

Please see: http://www.cprogramming.com/tutorial/bitwise_operators.html[^].

[EDIT]

But if the ultimate goal is to calculate the binary string representation, the solution is different. (Only not "decimal to binary": your integer values cannot be "decimal", only strings can be. You convert binary to binary strings, that is, extract bytes).

To find binary string, you need to extract each byte and fill in the string with 0 or 1. Here is how. First, create a bit mask, which should be a power of 2, see above. Do it in cycle for each bit. For a mask for each bit position, you have to find logical AND of the test number and the mask. The result is either zero or not. If zero, the text bit is clear, otherwise it is set:

C++

char bitDigit;
if ((1 << bit) & value) == 0)
    bitDigit = '0';
else
    bitDigit = '1';

Fill values of bitDigit left to right in the string, according to the bit position bit, in cycle.

That's all.

—SA

Posted 26-Jun-15 17:36pm

Sergey Alexandrovich Kryukov

Updated 27-Jun-15 2:12am

v3

Comments

nv3 27-Jun-15 1:19am

My 5!

Sergey Alexandrovich Kryukov 27-Jun-15 1:20am

Thank you.
—SA

nihal saxena 27-Jun-15 8:13am

so where should i place "int power = 1 << a;" at initialisation or between the loop

Sergey Alexandrovich Kryukov 27-Jun-15 8:14am

Please see my update to the question, after [EDIT].
—SA

Frankie-C 27-Jun-15 8:17am

5ed.
Numbers could be represented as a polynomial of powers of the base.
I.e. 13 = 1x10^1 + 3x10^0
When the base is 2 as in binary representation by shifting 1 is as moving a coefficient=1 to the polynomial value equal to the exponent.
So 1<<a is equivalent to 1x2^a

Sergey Alexandrovich Kryukov 27-Jun-15 8:20am

Thank you.
—SA

nihal saxena 27-Jun-15 10:41am

sir i am not getting the correct output and your i don't get your bitdigit concept so can u please do edit my code and paste the correct code -
#include<iostream>
using namespace std;
#include<conio.h>

int main()
{
int a,b,temp=0,n;
cout<<"Enter the range of natural no's : ";
cin>>b;
n=b;
for(a=0;a<=n;a++)
{
temp=temp+n%10;
cout<<"\nn : "<<n<<" temp : "<<temp<<" a : "<<a;
temp=temp*1<<a;
n=n/10;
}
cout<<"\nThe sum of given natural no's is : "<<temp;
getch();
return 0;
}

for example i have given the input 1001 and output coming is 8 but real output is 9

Sergey Alexandrovich Kryukov 27-Jun-15 13:32pm

This is because you are not doing what I advised. Calculation of power is not enough, it was just the answer to your immediate question. For next step, you need to follow my recipe I explained in my update above, after [EDIT].
—SA

nihal saxena 27-Jun-15 22:18pm

that's what i wanna ask how to use that code -
char bitDigit;
if ((1 << bit) & value) == 0)
bitDigit = '0';
else
bitDigit = '1';

Sergey Alexandrovich Kryukov 27-Jun-15 22:22pm

It's in the loop by bit.
What is unclear here? Did you try to use it?
—SA

nihal saxena 28-Jun-15 0:35am

no i dont get where to imply it in my code

Sergey Alexandrovich Kryukov 28-Jun-15 1:36am

"Where to imply" is a wrong way of thinking.
Did you get how to extract 1 bit from a value?
Then extract all bits in cycle. Don't ask "where". You write code, not anyone else. You decide "where".
—SA

nihal saxena 28-Jun-15 9:58am

no i dont know to extract 1 bit from value

Sergey Alexandrovich Kryukov 28-Jun-15 10:43am

I explained by my code sample, where "bit" is 0, 1, 2... — bit number, or bit position, counted from left to right (from least significant bit). Is it obvious why my Boolean expression find out if the bit is set or clear? If not, did you read the help page on bitwise operators?
—SA

nihal saxena 28-Jun-15 22:23pm

oh srry now i get it the thing is i have asked wrong i want to convert binary to decimal sorry

Sergey Alexandrovich Kryukov 28-Jun-15 22:51pm

(Sigh...) Let's start over. Write this code yourself and show what you have tried — in a separate question, if you fail to solve the problem. And I answered your current question; no matter if you asked it by mistake — the answer is still an answer — so please accept it formally.
—SA

nihal saxena 28-Jun-15 23:20pm

sorry again and i have accepted u r answer.

Sergey Alexandrovich Kryukov 29-Jun-15 1:28am

No problem. If you want me to participate in answering your next question, you can add another comment — to this comment. The problem of creating a decimal is even simpler.
Good luck.
—SA

nihal saxena 30-Jun-15 22:26pm

the tell me the way to decimal to binary, can it be done by modifying the above code

Sergey Alexandrovich Kryukov 30-Jun-15 22:54pm

Don't modify the code. Use the code I suggested in my answer.
Besides, you can also simply call Convert.ToString(value, 2);
—SA

Afzaal Ahmad Zeeshan 27-Jun-15 13:39pm

+5, interesting way of using bitwise operators.

Sergey Alexandrovich Kryukov 27-Jun-15 13:47pm

Thank you, Afzaal.

Interesting? Very usual, isn't it?

Here, I need to note this: if the goal is to get power, the knowledge of the bitwise representation of numbers is used. Apparently, bit shift won't get the power if used on the whole floating-point value, for example. If the goal is to extract bits, to show binary string representation, the bitwise approach is abstracted from the data representation and should always be used. Instead of power (arithmetic approach), it should be bitwise.

—SA

Afzaal Ahmad Zeeshan 27-Jun-15 14:38pm

Hmm, good advise for binary string representation. Thank you.

Sergey Alexandrovich Kryukov 27-Jun-15 15:02pm

My pleasure.
—SA