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savinki

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count digits

Jun 3rd, 2008

Hi,

wt is the function that can use to count number of digits in a long integer?
e.g. 12345678 -> 8
11 ->2
456 -> 3
6724 -> 4

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Duoas

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Re: count digits

Jun 4th, 2008

You'll need to use the modulo (or remainder) operator: %

It does what it says:
cout << (123 % 10);

Hope this helps.

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Re: count digits

Jun 4th, 2008

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Originally Posted by savinki

Hi,

wt is the function that can use to count number of digits in a long integer?
e.g. 12345678 -> 8
11 ->2
456 -> 3
6724 -> 4

I don't know that there is a function that does that. You can do two things though. You can use the log10 function from cmath:

http://www.cplusplus.com/reference/c...ath/log10.html

That function takes and returns doubles, so you'll need to do some typecasting and adjust by one.

Or you can set up a for-loop and divide by 10 each time through the loop using integer division. Count how many times you go through the loop.

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Re: count digits

Jun 4th, 2008

or....

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
int getNumDecimalDigits(int value)
{
    char myString[16];
    sprintf(myString,"%d",value);
    return strlen(myString);
}
int getNumDecimalDigits(int value)
{
    char myString[16];
    sprintf(myString,"%d",value);
    return strlen(myString);
}

Last edited by jephthah; Jun 4th, 2008 at 2:21 am.

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Re: count digits

-1

Jun 4th, 2008

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Originally Posted by savinki

Hi,

wt is the function that can use to count number of digits in a long integer?
e.g. 12345678 -> 8
11 ->2
456 -> 3
6724 -> 4

Hey Saviniki try this code:
#include<stdio.h>
void main()
{
long int a;
int count=0,;
printf("Enter a interger :");
scanf("%d",&a);
while(a>0)
{
a=a/10;
count++;
}
printf("No. of digits in the no entered : %d",count);
}

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Re: count digits

Jun 4th, 2008

@Deepak
Do not give away the solution. Read the last point of this link http://www.daniweb.com/forums/thread78223.html

Also posting C code in C++ forum is not a good idea that too without code tags.

Last edited by hammerhead; Jun 4th, 2008 at 4:37 am.

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Re: count digits

-1

Jun 4th, 2008

•

int getNumDecimalDigits(int value)
{
char myString[16];
sprintf(myString,"%d",value);
return strlen(myString);
}

That is not the best way to do it, heres a better way:

 Help with Code Tags 
 CPP Syntax (Toggle Plain Text) 
// Will work with negative numbers
int CountDidgets(int val) {
	int d = 1, c;
	if (val >= 0) for (c = 10; c <= val; c *= 10) d++;
	else for (c = -10 ; c >= val; c *= 10) d++;
	return (c < 0) ? ++d : d;
}
// Will work with negative numbers
int CountDidgets(int val) {
	int d = 1, c;
	if (val >= 0) for (c = 10; c <= val; c *= 10) d++;
	else for (c = -10 ; c >= val; c *= 10) d++;
	return (c < 0) ? ++d : d;
}

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Re: count digits

Jun 4th, 2008

> That is not the best way to do it
There's no such thing as the best overall way to do it. Your way might be better in Edward's program but worse in jephthah's, or vice versa. Though jephthah's function has the benefit of simplicity.

> heres a better way
The results are identical for the most part. Ed would argue that the sign has no business being counted as a digit, but that's something that depends on the needs of the program as well.

One big problem with your better way is that it fails to handle INT_MIN and INT_MAX.

Edward would do something more like this for the counting method:

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
int CountDigits(int value)
{
  int count = 0;
 
  value = std::abs(value);
 
  do {
    ++count;
    value /= 10;
  } while (value != 0);
 
  return count;
}
int CountDigits(int value)
{
  int count = 0;

  value = std::abs(value);

  do {
    ++count;
    value /= 10;
  } while (value != 0);

  return count;
}

Last edited by Radical Edward; Jun 4th, 2008 at 3:40 pm.

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Re: count digits

Jun 4th, 2008

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The results are identical

Not exactly, the function I made is Much more efficient and doesnt make any calls to any other functions. And I think the minus sign should count as a didget as if you are don't, trying to assign it to a char buffer and it wont have enough space if it's a negative number.

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Re: count digits

#10

Jun 4th, 2008

> the function I made is Much more efficient
Edward would argue that an infinite loop for some inputs isn't much more efficient.

Ed would also argue that any sane programmer will use a slow and correct function rather than a fast and broken function. Finally, pushing for efficiency at the cost of clarity without bottleneck benchmarks to back up the decision is usually a bad idea.

> trying to assign it to a char buffer and it wont have enough space if it's a negative number
That makes sense if all you use this function for is allocating buffers. There are other uses as well, and counting the sign would be treated as an error in many of them. I think you should avoid counting the sign and let the caller decide what to do:

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
int n = CountDigits(value);
 
if (value < 0 || includeSign)
  ++n;
 
...
int n = CountDigits(value);

if (value < 0 || includeSign)
  ++n;

...

But that's just a personal preference. As long as your choice is well documented, the caller can do the opposite just as easily:

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
int n = CountDigits(value);
 
if (value < 0)
  --n;
 
...
int n = CountDigits(value);

if (value < 0)
  --n;

...