Convert digits from string of numbers into ints then into array [duplicate]

Question

I have a 1000 digit number that I am storing as a string. I want to end up with each digit in that number as its own element in a vector. so "12345" would be {1,2,3,4,5}. Here is the code I have so far, but it is not working. The error is coming from comparing the string letter to the number. Any help would be appreciated. I have seen another question that is similar, but the string contains spaces between numbers, this doesn't.

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <sstream>
#include <string>

int main()
{


std::string input_string = "73167176531330624919225119674426574742355349194934"
"96983520312774506326239578318016984801869478851843"
"85861560789112949495459501737958331952853208805511"
"12540698747158523863050715693290963295227443043557"
"66896648950445244523161731856403098711121722383113"
"62229893423380308135336276614282806444486645238749"
"30358907296290491560440772390713810515859307960866"
"70172427121883998797908792274921901699720888093776"
"65727333001053367881220235421809751254540594752243"
"52584907711670556013604839586446706324415722155397"
"53697817977846174064955149290862569321978468622482"
"83972241375657056057490261407972968652414535100474"
"82166370484403199890008895243450658541227588666881"
"216427171479924442928230863465674813919123162824586"
"17866458359124566529476545682848912883142607690042"
"24219022671055626321111109370544217506941658960408"
"07198403850962455444362981230987879927244284909188"
"84580156166097919133875499200524063689912560717606"
"05886116467109405077541002256983155200055935729725"
"71636269561882670428252483600823257530420752963450";

std::cout << "storing string" << std::endl;

std::vector<int> return_vector;

for (int i = 0; i<1000; i++)
{
    if (&input_string[i] == "0")
    {
        std::cout << 0 << std::endl;
        return_vector.push_back(0);
    }

    if (&input_string[i] == "1")
    {
        return_vector.push_back(1);
    }

    if (&input_string[i] == "2")
    {
        return_vector.push_back(2);
    }

    if (&input_string[i] == "3")
    {
        return_vector.push_back(3);
    }

    if (&input_string[i] == "4")
    {
        return_vector.push_back(4);
    }

    if (&input_string[i] == "5")
    {
        return_vector.push_back(5);
    }

    if (&input_string[i] == "6")
    {
        return_vector.push_back(6);
    }

    if (&input_string[i] == "7")
    {
        std::cout << 7 << std::endl;
        return_vector.push_back(7);
    }

    if (&input_string[i] == "8")
    {
        return_vector.push_back(8);
    }

    if (&input_string[i] == "9")
    {
        return_vector.push_back(9);
    }




}

std::cout << "processing string" << std::endl;


for (int j = 0; j<1000; j++)
{
    std::cout << return_vector[j] << std::endl;
}

std::getchar();
return 0;
}

Replace if (&input_string[i] == "0") with if (input_string[i] == '0') and so on. And there are infinitely simpler ways to do this... — BlackDwarf
– BlackDwarf, Commented Oct 8, 2015 at 15:17
The error is coming from comparing the string letter to the number Tell us the exact error message. — eerorika
– eerorika, Commented Oct 8, 2015 at 15:19

ashiquzzaman33 · Accepted Answer · 2015-10-08 15:50:18Z

1

To compare char in string use

input_string[i] == '0'

So replace

if (&input_string[i] == "0")

with

if (input_string[i] == '0')  //Removed & from &input_string[i], and replace double quotation(") to single quotation(').

Better Approach

You don't need bulk size of if-else, you can use like

return_vector.reserve(input_string.length());  //Requests  vector capacity be at least enough to contain input_string.length() number of element, which reduce redundant reallocation.

for (int i = 0; i<input_string.length(); i++)
{
    return_vector.push_back(input_string[i]-'0');
}

edited Oct 8, 2015 at 15:50

answered Oct 8, 2015 at 15:18

ashiquzzaman33

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2 Comments

eerorika Over a year ago

@JoshJackson this does work because the types are different.

LogicStuff Over a year ago

Too much reallocations without std::vector::reserve.

NathanOliver · Accepted Answer · 2015-10-08 15:29:07Z

1

We can use std::transform and a lambda to do this

std::string test = "123456789";
std::vector<int> result(test.size());
std::transform(test.begin(), test.end(), result.begin(), [](char ch) { return ch - '0'; });

Live Example

edited Oct 8, 2015 at 15:29

answered Oct 8, 2015 at 15:21

NathanOliver

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Comments

Tony Delroy · Accepted Answer · 2015-10-08 15:21:53Z

0

You can convert mathematically between ASCII character codes and numbers - just subtract '0' (which is 48 decimal):

for (char c : input_string)
    return_vector.push_back(c - '0');

answered Oct 8, 2015 at 15:21

Tony Delroy

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Comments

Semyon Tikhonenko · Accepted Answer · 2015-10-08 15:27:12Z

0

Try this:

   std::vector<int> toVector(std::string str)
    {
        std::vector<int> result(str.size());
        std::transform(str.begin(), str.end(), result.begin(), [](char ch){return ch-'0';});
        return result;
    }

edited Oct 8, 2015 at 15:27

answered Oct 8, 2015 at 15:24

Semyon Tikhonenko

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