Remove digits in number

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 :)
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you can use shift operator
@kulkarnisr : How?
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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.
@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.
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Well then, add some logic to determine how big the number is first.
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.
 ``12345678910111213`` ``````#include #include 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
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@mario0815

If you do log10(num)+1, you get the number of digits in the number...
 ``123456`` ``````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 ; }``````
 ``123456`` ``````unsigned remove_leftmost_digit(unsigned n) { static char temp[11]; sprintf(temp, "%d", n); return atoi(temp+1); }``````

 ``12345678910`` ``````unsigned remove_leftmost_digit(unsigned n) { // static char temp[11]; static char temp[ std::numeric_limits::digits10 + 2 ] ; // sprintf(temp, "%d", n); sprintf(temp, "%u", n) ; return atoi(temp+1); }``````
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?
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oops..! sorry ziv...
@abhishek thanks
> 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`
ok.. thanks JLBorges!
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