Remove digits in number

[ZivHirsch]

I know how to remove digits in number from right to left.
For example: the number 319. If I do (number /= 10), I get 31.

My question is how can I remove digits in number from left to right.
For example: the number 319. If I will do something, I will get the number 19.

Thank you :)

[kulkarnisr]

you can use shift operator

[ZivHirsch]

@kulkarnisr : How?

[abhishekm71]

Shift operator works only on the binary level, i.e., divide by 2, multiply by 2 etc.

For the OP's requirement % (remainder) operator is suitable.
319 % 1000 = 319.
319 % 100 = 19.
319 % 10 = 9.

[ZivHirsch]

@abhishekm71 : I gave that number only for example, I dont pick the number, so I dont know how to devide it.
If the number is 168735
and I do (168735 % 100) I will get 35, I wont get all the numbers besides 1 from the left like u did.

I hope u understand me.

[MikeyBoy]

Well then, add some logic to determine how big the number is first.

[abhishekm71]

Then perhaps you will either have to count the number of digits in the number (using a loop) and apply % accordingly, or take input as a string if you can, modify it as wished and use stringstream to convert it into decimal whenever required.

[mario0815]

#include <iostream>
#include <math.h>

using namespace std;

int DropLeadingDigit(int number){return(number % (int) pow((double) 10, (double) floor(log((double) number) / log((double) 10))));}

int main(){
    int number;
    while(cin >> number) cout << DropLeadingDigit(number) << endl << endl;
    
    return(0);
}

floor(log(a) / log(b)) + 1 is the number of digits of a in base b

[Smac89]

@mario0815

If you do log10(num)+1, you get the number of digits in the number...

[JLBorges]

unsigned int remove_most_significant_digit( unsigned int n )
{
    int divisor = 1 ;
    for( int v = n ; v > 9 ; v /= 10 ) divisor *= 10 ;
    return n % divisor ;
}

[tntxtnt]

unsigned remove_leftmost_digit(unsigned n)
{
    static char temp[11];
    sprintf(temp, "%d", n);
    return atoi(temp+1);
}

[JLBorges]

unsigned remove_leftmost_digit(unsigned n)
{
    // static char temp[11];
    static char temp[ std::numeric_limits<unsigned int>::digits10 + 2 ] ;

    // sprintf(temp, "%d", n);
    sprintf(temp, "%u", n) ;

    return atoi(temp+1);
}

[abhishekm71]

why std::numeric_limits<unsigned int>::digits10 + 2 ?
why not std::numeric_limits<unsigned int>::digits10 + 1 ?

EDIT: I know that '\0' occupies 1 extra location. Any other character that is needed?

[kulkarnisr]

oops..! sorry ziv...
@abhishek thanks

[JLBorges]

> I know that '\0' occupies 1 extra location. Any other character that is needed?

The number of decimal digits in std::numeric_limits<unsigned int>::max() is one greater than the value of std::numeric_limits<unsigned int>::digits10

[abhishekm71]

ok.. thanks JLBorges!