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reading number

Posted on 2004-10-15
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Last Modified: 2010-04-01
Just a quick question.

How do we convert 10,000 to 10000?
I read the number using char array from a file, and want to copy thearray to an integer variable. Is that possible?
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Question by:icysmarty
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Expert Comment

by:Mysidia
Comment Utility
Just the same way you convert "10000" to int representation 10000
if you don't use library calls.

like this:

while(x < array_size)
  num = num * 10;
  num = num + array[x];
  x = x + 1;
}

just add checks to skip commas.
x and num  start at zero
and 'array_size' should be the actual length of the string
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Expert Comment

by:Mysidia
Comment Utility
er,

num = num + (array[x]-'0')

or if you prefer

num = num + charToDigit(array[x])


int charToDigit(char v)
{
  switch(v) {
     default: return 0; break;
     case '0': return 0;
     case '1': return 1;
     case '2': return 2;
     case '3': return 3;
...
  }
}
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Expert Comment

by:RogueAce
Comment Utility
What's wrong with library calls?

#include <cstdlib> to get atoi and <ctypes> to get isdigit, if I recall correctly.

num=0;
for(unsigned int c=0; c < arraysize; c++){
  if( isdigit( myarray[c] )){
    num*=10;
    num += atoi(myarray[c]);
  }
}
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Expert Comment

by:Mysidia
Comment Utility
The fact that atoi takes a pointer to a string as its parameter,
and that segment passes an invalid input to atoi()

Moreover, building a string just to use atoi() would be
wasteful, and there's no reason to go through all the
trouble involved.

int num = 0, x = 0;

while(x < array_size)
  if (!isdigit(x)) {
      /* Detect, skip invalid characters,  report such errors as per the needs
          of your situation*/
      continue;
  }
  num = num * 10;
  num = num + (array[x] - '0');
  x = x + 1;
}
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Author Comment

by:icysmarty
Comment Utility
Does this apply to 10,000 only? What if I have 12,345 and 56,123?
I am doing this

void read_byte( unsigned char byte)
{
    ....
    ....
    if(byte == 44)
    {}
    else
    array[i++] = byte;
}

The problem is I dunno how to make it display as an integer say 12345, instead of digit 1 digit 2 digit 3 digit 4 digit 5.
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Accepted Solution

by:
Mysidia earned 200 total points
Comment Utility
The answer is that the strategy works for an arbitrary positive integer inside the legitimate range for
integers.

It can trivially be extended to negative integers by handing the -.

Consider input digits
     D0   D1  D2  D3   D4

What does it mean for a digit to be in the  position on the far right?
Why it means that it is multiplied by 10^0

Second position is weighted as 10^1,  Third position from the right
is 10^2, etc.

Ok, What's num equal to after reading all four example digits then?

Well consider Num_[NNN] as a representation of num after the NNNth iterate

The first iterate of the loop is 0  what is Num after that?
Why:
Num_0 = 0*0 + D0

Guess what, it is in general true that for the algorithm:
Num_(i+1) = (Num_i * 10) + D_i+1

And:
Num_1 = Num_0 * 10 + D1 = D0 * 10 + D1
Num_2 = Num_1*10 + D2 = D0*100 + D1*10 + D2
Num_3 = Num_2*10 + D3 = D0*1000 + D1*100 + D2*10 + D3
Num_4 = Num_3*10 + D3 = D0*10000 + D1*1000 + D2*100 + D3*10 + D4

So if you let the four input digits in the example be
                D1 =  5
                D2  = 4
                D3  = 3
                D4   = 2

then the result is

num = 5*1000 + 4*100 + 3*10 + 2


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Expert Comment

by:Mysidia
Comment Utility
legitimate range of integers can be determined by #includ'ing  <limits.h>

the maximum value is INT_MAX
the minimum value is called INT_MIN
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Expert Comment

by:Mysidia
Comment Utility
Err, oops technically:
 Num_(i+1) = (Num_i * 10) + D_i+1

should be more like
   Num_(i+1) = (Num_(i-1) * 10) + D_i
with initial condition  Num_(-1) = 0        [Since D0 is the first numbered digit]
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Author Comment

by:icysmarty
Comment Utility
I got the concept and has coded it according to the logic. Thank you.
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Expert Comment

by:RogueAce
Comment Utility
Hmm, you're right about atoi. I forgot it took a c-style string instead of a character.

The short version of what Mysidia is saying, icysmarty, is that it will work for any reasonably-sized (i.e. under two billion) number.
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