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Jun 18th, 2009

count digits in string

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C++ Syntax (Toggle Plain Text)
int countDigits(string word)
{
    int count = 0;
 
    for (int i = 0; i < word.size(); i++)
    {    
        if (word.at(i) < 57 && word.at(i) > 49)
        {
            count++;
        }
    }
}
int countDigits(string word)
{
    int count = 0;
    
    for (int i = 0; i < word.size(); i++)
    {    
        if (word.at(i) < 57 && word.at(i) > 49)
        {
            count++;
        }
    }
}

i want to count the digits in a string "abc123". i know my problem is in the if statemnt i just cant get it working

stupid question i know im just stuck

Last edited by timb89; Jun 18th, 2009 at 11:43 pm.

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timb89

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Jun 19th, 2009

Re: count digits in string

you can use the macro isdigit() to find out if it is '0' to '9'. And the At() method is unnecessary.

if( isdigit(word[i]) )

Last edited by Ancient Dragon; Jun 19th, 2009 at 12:18 am.

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Jun 19th, 2009

Re: count digits in string

Click to Expand / Collapse Quote originally posted by Ancient Dragon ...

you can use the macro isdigit() to find out if it is '0' to '9'. And the At() method is unnecessary.

if( isdigit(word[i]) )

Is isdigit a macro then?
According to this information it isn't

tux4life

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Jun 19th, 2009

Re: count digits in string

Change your comparison to
if (word[i] < 57 && word[i] > 49)

Also, the line is better written as
if (word[i] > 49 && word[i] < 57)
It's a more logical order for the comparison pair.

WaltP

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Jun 19th, 2009

Re: count digits in string

Click to Expand / Collapse Quote originally posted by WaltP ...

Change your comparison to
if (word[i] < 57 && word[i] > 49)

Also, the line is better written as
if (word[i] > 49 && word[i] < 57)
It's a more logical order for the comparison pair.

Yes, but isn't the isdigit function a bit easier to work with?
However if he isn't allowed to use that function it would seem more logical to me if he'd written it like you did

Last edited by tux4life; Jun 19th, 2009 at 3:30 am.

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Jun 19th, 2009

Re: count digits in string

And why 49 and 57 instead of '0' and '9'??? Are you training for an obfuscation contest?

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Jun 19th, 2009

Re: count digits in string

Click to Expand / Collapse Quote originally posted by jencas ...

And why 49 and 57 instead of '0' and '9'??? Are you training for an obfuscation contest?

Agreed, '0' and '9' makes your code more readable, but the OP started with using the ASCII codes

But this is actually not a good reason to avoid this I think

Last edited by tux4life; Jun 19th, 2009 at 4:01 am.

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Jun 19th, 2009

Re: count digits in string

I would solve it in this efficient easy way

C++ Syntax (Toggle Plain Text)
#include <iostream>
#include <string>
using namespace std;
 
int countdigits(string word)
{
	int count=0;
	for(int i=0;i<(int)word.length();i++)
	{
		if(isdigit(word[i]))
			count=count+1;
	}
	return count;
}
void main()
{
	string word;
	getline(cin,word);
	int digits=0;
	digits=countdigits(word);
	cout<< digits<<endl;
}
#include <iostream>
#include <string>
using namespace std;

int countdigits(string word)
{
	int count=0;
	for(int i=0;i<(int)word.length();i++)
	{
		if(isdigit(word[i]))
			count=count+1;
	}
	return count;
}
void main()
{
	string word;
	getline(cin,word);
	int digits=0;
	digits=countdigits(word);
	cout<< digits<<endl;
}

Last edited by Ancient Dragon; Jun 19th, 2009 at 10:19 am. Reason: add code tags, remove color and bold tags

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Jun 19th, 2009

Re: count digits in string

Click to Expand / Collapse Quote originally posted by tux4life ...

Agreed, '0' and '9' makes your code more readable, but the OP started with using the ASCII codes
But this is actually not a good reason to avoid this I think

I didn't quote your answer because my answer was intended to address the TO.

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Jun 19th, 2009

Re: count digits in string

Click to Expand / Collapse Quote originally posted by tux4life ...

Is isdigit a macro then?
According to this information it isn't

Its probably implementation dependent. You have to check your compiler's ctype.h header file to find out whether its implemented as a macro or a function. The link you posted doesn't say one way or the other.

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