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?

Vittorio Romeo · Accepted Answer · 2018-09-13 13:40:10Z

5

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.

answered Sep 13, 2018 at 13:40

Vittorio Romeo

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1 Comment

teekanne Over a year ago

Thank you for your answer. It helped me to understand my error and find a first bugfix.

Johnny Johansson · Accepted Answer · 2018-09-13 13:51:09Z

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();
}

answered Sep 13, 2018 at 13:51

Johnny Johansson

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1 Comment

teekanne Over a year ago

Thank you for your solution. I've got another solution with the help of @VittorioRomeo at first, but yours is way more elegant. I use your solution now

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