如何修剪 32 位二进制数的所有前面的零？

Question

I was trying to trim all the leading/preceding zeroes in a binary 32-bit number and print out the output by these following 2 codes of which the first one showed error while the other code was successful:

I want to know why the former one failed. please explain in detail. it would mean a lot to me.

    string trimmer(string& binary_str) {

    int i=0;// 'i' will finally have the index of the binary string which is having value as '0' and is 
            //  just the preceding zero to the first 1 in the string 

    while (binary_str.at(i) != 1) {//this will iterate over the srting until it reaches the zero before 
                                   //the first 1 in the string
        i++;
    }
    binary_str.erase(0, i);      //this will erase all the preceeding zeroes
    return binary_str;
}

This above written code didnt work and showed up error,

But this second code worked:

string trimmer(string& binary_str) {
    int i=0;
    for (i = 0; i < 32; i++) {
        if (binary_str.at(i) == '1') {
            break;
        }
    }
    binary_str.erase(0, i);
    return binary_str;
}

Both had the function, ie, if input was given: a binary string 00000000000000000000001011001011 then the output should be 1011001011.

The former code used while loop whereas the latter used for loop with if conditional. bot codes follow the same logic. but the former one with while loop showed up error whereas the latter one didn't.

Can anybody explain to me the reason behind the while loop to show up error? it would mean a lot to me.

The complete code is as follows:

#include<bits/stdc++.h>
using namespace std;
/*
string trimmer(string& binary_str) {
    int i=0;
    while (binary_str.at(i) != 1) {
        i++;
    }
    binary_str.erase(0, i);
    return binary_str;
}
*/
string trimmer(string& binary_str) {
    int i=0;
    for (i = 0; i < 32; i++) {
        if (binary_str.at(i) == '1') {
            break;
        }
    }
    binary_str.erase(0, i);
    return binary_str;
}

int main(){
    string bStr="00000000001011001011"; 
    bStr = trimmer(bStr);
    cout<< bStr<<'\n';
    return 0;
}

Answer 1

You can use find_first_not_of to find the first character of that isn't a 0 . Then you can erase all characters up to that point. If a string is all zeros, it will be completely erased.

void trim_leading_zeros(std::string &str) {
    str.erase(0, str.find_first_not_of('0'));
}

Answer 2

Try using modern C++ features:

#include <iostream>
#include <algorithm>
#include <string>

void ltrim_zero(std::string& str) {
    str.erase(str.begin(), std::find_if(str.begin(), str.end(), [](int character) {
        return '0' != character;
    }));
}

int main(){
    std::string bStr = "00000000001011001011"; 
    ltrim_zero(bStr);
    std::cout << bStr << std::endl;
    return 0;
}

Try live

Answer 3

find the first occurrence of '1' and use this variation of string constructor

    string (const string& str, size_t pos, size_t len = npos);

str=string to be copied from

pos= starting position to copy from

npos= number of character to copy from source(str) string.