Solved

Posted on 2004-10-15
218 Views
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?
0
Question by:icysmarty
• 6
• 2
• 2

LVL 23

Expert Comment

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
0

LVL 23

Expert Comment

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;
...
}
}
0

LVL 1

Expert Comment

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]);
}
}
0

LVL 23

Expert Comment

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
continue;
}
num = num * 10;
num = num + (array[x] - '0');
x = x + 1;
}
0

Author Comment

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

{
....
....
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.
0

LVL 23

Accepted Solution

Mysidia earned 200 total points
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

0

LVL 23

Expert Comment

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
0

LVL 23

Expert Comment

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]
0

Author Comment

I got the concept and has coded it according to the logic. Thank you.
0

LVL 1

Expert Comment

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.
0

## Featured Post

### Suggested Solutions

C++ Properties One feature missing from standard C++ that you will find in many other Object Oriented Programming languages is something called a Property (http://www.experts-exchange.com/Programming/Languages/CPP/A_3912-Object-Properties-in-C.ht…
This article shows you how to optimize memory allocations in C++ using placement new. Applicable especially to usecases dealing with creation of large number of objects. A brief on problem: Lets take example problem for simplicity: - I have a G…
The viewer will learn how to user default arguments when defining functions. This method of defining functions will be contrasted with the non-default-argument of defining functions.
The viewer will learn how to clear a vector as well as how to detect empty vectors in C++.