How to create 'char *[]' in c++?

Question

I am trying to call into a C library from C++. The interface requires char* [] . The compilation fails if I pass const char * [] .

How do I safely create one of these? This is the best I can do, but I don't know if it's even correct. I'm creating the array with 1 c-string in it only.

string attr = "memberUid";
vector<char> attr_v(attr.begin(), attr.end());
attr_v.push_back('\0');
char * attrs[] = {&attr_v[0]};

I do not want to trigger the deprecated conversion warning, ie the one that fires with the following:

char * attr = "memberUid";

Answer 1

Assuming you have a vector of strings ( std::vector<std::string> ), which I'll call source :

std::vector<char*> asArgument( source.size() );
std::transform(
    source.begin(), source.end(),
    asArgument.begin(),
    []( std::string const& elem ) { return const_cast<char*>( elem.c_str() ); } );
cFunction( asArgument.data(), asArgument.size() );

If you need the array of char* to be NULL terminated, just create asArgument with source.size() + 1 .

Of course, if you need string literals:

char const* asArgument[] = { ... };

will do the trick, with a const_cast at the call site ( const_cast<char**>( asArgument ) ).

This assumes that the called function doesn't actually try to modify anything; that the absence of const is only because it is legacy code. If it does modify any of the strings passed to it, you'll have to make deep copies.

Answer 2

hope this will help , for loop is need to init the whole array

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main()
{
  vector<string> attr_v;
  attr_v.push_back("10");
  char * attr[100];
  attr[0]= const_cast<char*>(attr_v[0].c_str()); 
   return 0;
}