如何在C ++ execve命令中将字符串转换为char [] *？

Question

When trying to use the execve command, I receive an error about my last 2 arguments.

     #include <unistd.h>        
     const char * c = enviorment.c_str();
     execve(full.c_str() , cl.getArgVector(), c);

I've tried using the several different options in copying strings such as:

char *temp2 = new char[(path).size()+1];
strcpy ( temp2, cl.getCommand());
    execve(full.c_str() , cl.getArgVector(), temp);

but I'm unable to get it to work and recieve an error message:

error: cannot convert char* to char* const* for argument 3 to int execve(const char*, char* const*, char* const*)

Answer 1

The environment variable array is an array of strings (char*), not one string. It needs a NULL value as the final element in the array.

Mimic the code you're using to implement c1.getArgVector() making sure the last char* in the array is NULL.