What is wrong with this code? I need to convert string into float

Question

What is wrong with this code? I need to convert string into float. m2 has an error in m2.length

#include <iostream>
#include <string>
#include <cstdlib>
#include <sstream>
using namespace std;

int main()
{
    float v2;
    char *m2 = "1 23 45 6";
    for (int i = 0; i < m2.length(); i++) //to convert every element in m2 into float
    {
        v2 = atof(&m2[i]);
    }
    printf("%.2f", v2);

    system("pause");
    return 0; 
}

Answer 1

I need to convert each element in the array so that I can store them to make operations

Okay so how about you use a string stream to extract the numbers from the string into a vector?

#include <iostream>
#include <iterator>
#include <sstream>
#include <vector>

int main()
{
    std::string input = "1 23 45 6";
    std::stringstream ss(input);
    std::istream_iterator<double> end;
    std::vector<double> output(std::istream_iterator<double>(ss), end);
}

Answer 2

There are too may things wrong with your code to explain each and every point. That's how I would do it:

float v2;
std::istringstream iss("1 23 45 6");
while(iss >> v2) {
    std::cout << v2 << std::endl;
}

Answer 3

Step by step analysis, for learning from the errors:

First there is no member function length() for an array of characters, so you should define string m2="..."; in order for the code to compile.

Then unfortunately your code would print only one number: the last. Why ? Because your printf() is outside the loop.

Once you've put it inside the loop, you'd have plenty of numbers, but more than you expect:
* the first iteration, it'd start with "1 23 45 6" => 1
* the second with " 23 45 6" => 23
* the third iteraton "23 45 6" => 23 again!
* the fourth iteration "3 45 6" => 3 (did you expect this ?)

So you'd have to jump from one number to the next. So instead of incrementing, you coud search for the next space, with function find_first_of():

for (int i = 0; i < m2.length(); i=m2.find_first_of(' ', i+1)) // jump to the next space
{
    v2 = atof(&m2[i]);
    printf("%.2f\n", v2);
}

Here the online demo.

Real c++ alternatives:

Have a look at πάντα ῥεῖ 's solution: that's the way we do it in C++

Answer 4

Use string instead of char*, to make things a little easier.

For converting a given string to a numerical value, either use stringstream or atof().

Quick solution to your problem:

#include <iostream>
#include <string>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
using namespace std;

int main()
{
    vector<float> myFloats;
    string m("1. 2.23 45 6456    72.8  46..");

    // First of parse given string and separate char groups that serve as a number
    // => only consider 0-9 and '.'
    if(m.size() == 0)
        return 1;

    string charGroup;
    for(int i=0; i<m.length(); i++)
    {
        if((isdigit(m.at(i)) || m.at(i) == '.'))
        {
            charGroup += m.at(i);
            if(i == m.length() - 1 && !charGroup.empty())
            {
                // in case given group is a numerical value
                myFloats.push_back((float) atof(charGroup.c_str()));

                // prepare for next group
                charGroup.clear();
            }
        }
        else if(!charGroup.empty())
        {
            if(m.at(i) == ' ')
            {
                // in case given group is a numerical value
                myFloats.push_back((float) atof(charGroup.c_str()));

                // prepare for next group
                charGroup.clear();
            }
            else charGroup.clear();
        }
    }

    // Print float values here
    if(!myFloats.empty())
    {
        cout << "Floats: ";
        for(int i=0; i<myFloats.size(); i++)
            cout << myFloats.at(i) << ", ";
    }

    getchar();
    return 0; 
}