如何使用 C++ 从字符串中删除前导零？

Question

I want to remove leading zeroes from string like "0000000057" .

I did like this but didn't get any result:

string AccProcPeriNum = strCustData.substr(pos, 13);

string p48Z03 = AccProcPeriNum.substr(3, 10);

I want output only 57 .

Any idea in C++?

Answer 1

#include <string>    

std::string str = "0000000057";
str.erase(0, str.find_first_not_of('0'));

assert(str == "57");

LIVE DEMO

Answer 2

Piotr S's answer is good but there is one case it will return the wrong answer, that is the all-zero case:

000000000000

To consider this in, use:

str.erase(0, min(str.find_first_not_of('0'), str.size()-1));

Even when str.size() will be 0 , it will also work.

Answer 3

Although it's probably not the most efficient (in terms of run-time speed) way to do things, I'd be tempted to just do something like:

std::cout << std::stoi(strCustData);

This is simple and straightforward to use, and gives a reasonable output (a single '0') when the input consists entirely of 0 s. Only when/if profiling showed that this simple solution was a problem would I consider writing substantially more complex code in the hopes of improving speed (and I doubt that would arise).

The obvious limitation here is that this does assume that the characters after the leading zeros are digits. That's clearly true in your sample, but I suppose it's conceivable that you have data where it's not true.

Answer 4

This should work as a generic function that can be applied to any of the std::basic_string types (including std::string ):

template <typename T_STR, typename T_CHAR>
T_STR remove_leading(T_STR const & str, T_CHAR c)
{
    auto end = str.end();

    for (auto i = str.begin(); i != end; ++i) {
        if (*i != c) {
            return T_STR(i, end);
        }
    }

    // All characters were leading or the string is empty.
    return T_STR();
}

In your case you would use it like this:

string x = "0000000057";
string trimmed = remove_leading(x, '0');

// trimmed is now "57"

( See a demo .)

Answer 5

#include<algorithm>
#include<iostream>
#include<string>
using namespace std;


int main()
{
   string s="00000001";
   cout<<s<<endl;
   int a=stoi(s);
   cout<<a;
   //1 is ur ans zeros are removed


}

Answer 6

This C language friendly code removes leading zeros and can be used as long as the char array's length fits within an int :

char charArray[6] = "000342";
int inputLen = 6;

void stripLeadingZeros() {
    int firstNonZeroDigitAt=0, inputLen = strlen(charArray);

    //find location of first non-zero digit, if any
    while(charArray[firstNonZeroDigitAt] == '0')
        firstNonZeroDigitAt++;

    //string has no leading zeros
    if(firstNonZeroDigitAt==0)
        return; 

    // string is all zeros
    if(firstNonZeroDigitAt==inputLen) {
        memset(charArray, 0, sizeof charArray);
        strcpy(charArray, "0");
        inputLen = 1;
        return;
    }

    //shift the non-zero characters down
    for(int j=0; j<inputLen-firstNonZeroDigitAt; j++) {
        charArray[j] = charArray[firstNonZeroDigitAt+j];
    }

    //clear the occupied area and update string length
    memset(charArray+inputLen-firstNonZeroDigitAt, 0, inputLen-firstNonZeroDigitAt+1);
    inputLen -= firstNonZeroDigitAt;
}