C++: convert vector of floats to char*

Question

How can I convert vector of floats into a char*?

I have a collection of floats stored inside std::vector<float> id_list :

0,
0,
0,
0,
0,
1.77636e-15,
2.35099e-38,
-7.10543e-15,
3.06107e-38,
....

and using this code to convert it:

char *ptr = (char*)&id_list[0];
std::string dump_string(ptr, id_list.size() * sizeof(float));

but dump_string.c_str() returns empty, despite having some values stored in my vector. I'm expecting to get all values concatenated into a one, long string, ie.:

0,0,0,0,0,1.77636e-15,2.35099e-38,-7.10543e-15,3.06107e-38........

Answer 1

I'm expecting to get all values concatenated into a one, long string, ie.:

0,0,0,0,0,1.77636e-15,2.35099e-38,-7.10543e-15,3.06107e-38........

You are creating a std::string that simply holds a copy of the raw bytes of the float array. So, if any of the bytes happen to contain a numeric value of 0 (as is the case in your example), you will get a truncated result if you try to treat the std::string as a null-terminated C string .

If you want to actually convert the float values into a human-readable string, you need to format the values individually, such as with std::ostringstream or std::to_string() , eg:

std::ostringstream oss;
if (!id_list.empty()) {
    oss << id_list[0];
    for(size_t i = 1; i < id_list.size(); ++i) {
        oss << "," << id_list[i];
    }
}
std::string dump_string = oss.str();

std::string dump_string;
if (!id_list.empty()) {
    dump_string += std::to_string(id_list[0]);
    for(size_t i = 1; i < id_list.size(); ++i) {
        dump_string += ',';
        dump_string += std::to_string(id_list[i]);
    }
}

Answer 2

. I'm expecting to get all values concatenated into a one, long string, ie.:

You could write a small function to do that.

Example:

#include <iostream>
#include <iterator> // std::advance
#include <string>
#include <sstream>  // std::stringstream
#include <vector>

// a function taking two iterators and streaming the content to a stringstream
// then returning a `std::string` using `stringstream::str()`
template<typename Begin, typename End = Begin>
std::string join(Begin begin, End end) {
    std::stringstream retval;
    if(begin != end) {
        retval << *begin;
        for(std::advance(begin, 1); begin != end; std::advance(begin, 1)) {
            retval << ',' << *begin;
        }
    }
    return retval.str();
}

int main()
{
    std::vector<float> id_list {
        0,
        0,
        0,
        0,
        0,
        1.77636e-15,
        2.35099e-38,
        -7.10543e-15,
        3.06107e-38,
    };
    std::cout << join(id_list.begin(), id_list.end());
}

Output:

0,0,0,0,0,1.77636e-15,2.35099e-38,-7.10543e-15,3.06107e-38

Answer 3

As others note, your approach of using casting is not the right way to go. Also, note that all of the value in your example are probably exactly 0.0 (or -0.0)! Why? Because they're beyond the precision range of float on typical platforms.

Still, you could concatenate the string representation of the float 's in an std::vector , using the ranges library , and with no need for any loops:

#include <vector>
#include <string>
#include <iostream>
#include <range/v3/all.hpp>

std::string concat_my_floats(const std::vector<float>& vec) 
{
    std::ostringstream oss;
    auto convert = [&](float x) { oss.str(""); oss << x; return oss.str(); };
    return vec  
        | ranges::views::transform(convert) 
        | ranges::views::cache1 // We need this, see footnote.
        | ranges::views::join(',')
        | ranges::to<std::string>();
}

int main () {
    std::vector<float> id_list = {0, 1.1, -2.2, 3.3};
    std::cout << concat_my_floats(id_list) << std::endl;
}

This will give you:

0,1.1,-2.2,3.3

---

_{If you're wondering what's that cache1 business - it has to do with how the transformed range is a range of prvalues, which the ranges library is not willing to join for performance reasons; apparently you need to explicitly allow a caching of the last element, expensive though it may be. See here .}

Answer 4

How can I convert vector of floats into a char*?

Here is one way to reinterpret the array of floating point numbers as bytes:

char *ptr = reinterpret_cast<char*>(id_list.data());

Which does exactly what your cast did except it doesn't rely on a C style cast.

---

but dump_string.c_str() returns empty

This is because dump_string happens to contain an empty null terminated string at the beginning.

---

I'm expecting to get all values concatenated into a one, long string, ie.:

 0,0,0,0,0,1.77636e-15,2.35099e-38,-7.10543e-15,3.06107e-38........

You expectation is misguided. Reinterpreting bytes of floating point numbers is a separate thing from serialising them to readable text. If this is the output you want , then reinterpretation is not the operation that you need.

You can use for example a string stream to convert floating point numbers to text.