Initialize a const char* Array

Question

I'm working in C++17 project, but have to use a C-legacy-library. For this I have to create a const char* array in C-style, but im struggling with the initialization. In particular,

#include <iostream>

int main(int argc, char* argv[])
{
    const int bound = 3;

    const char* inames[bound]{ "" };
    for(unsigned int i = 0; i < bound; i++) {
        const char *number = std::to_string(i).c_str();
        inames[i] = number;
    }

    for(unsigned int i = 0; i < bound; i++) {
        std::cout << "inames["
                  << std::to_string(i)
                  << "] = "
                  << inames[i]
                  << std::endl;
    }

    return 0;
}

returns

inames[0] = 2
inames[1] = 2
inames[2] = 2

as output, which I don't unterstand. I expected the output to be

inames[0] = 0
inames[1] = 1
inames[2] = 2

Can anyone please help me point me to my error?

Answer 1

You code has undefined behavior .

std::to_string(i).c_str()

You are creating a temporary std::string instance, then getting its internal const char* pointer. At the end of the line the temporary instance is dead, so the pointer is now dangling.

Answer 2

The problem is you don't have anywhere to actually store the strings themselves, only the pointers to them.

By doing it this way, the strings are stored in std::strings while beeing referenced by the plain C array:

const int bound = 3;
std::vector<std::string> strings(bound);
const char* inames[bound];
for (unsigned int i = 0; i < bound; i++) {
    strings[i] =  std::to_string(i);
    inames[i] = strings[i].c_str();
}