So I want to pass arguments to following function
execve(char *filename, char *argv[], char *envp[])
currently my argv[] is a string array. I want to convert it into a char* array so I can pass it to this function.
I have looked around and found many ways to convert a string to a char array but how to convert a string array to an array of char array I guess would be the right term
Any help?
You'll need to get the address of the data inside your std::strings. Note, that these are not required to be null-terminated, i.e., you'll need to make sure that all strings are null terminated. Also, the array passed as argv also needs to have the last element to be a null pointer. You could use code along the lines of this:
std::string array[] = { "s1", "s2" };
std::vector<char*> vec;
std::transform(std::begin(array), std::end(array),
std::back_inserter(vec),
[](std::string& s){ s.push_back(0); return &s[0]; });
vec.push_back(nullptr);
char** carray = vec.data();
When compiling with C++03, there are some changes necessary:
nullptr you need to use 0.std::string is not guaranteed to be contiguous, i.e., you need an additional, auxiliary std::vector<char> to hold a contiguous sequence of characters.There are no functions begin() and end() deducing the size of an array but they can easily be implemented in C++03:
template <typename T, int Size> T* begin(T (&array)[Size]) { return array; }
template <typename T, int Size> T* end(T (&array)[Size]) { return array + Size; }
The C++03 std::vector<T> doesn't have a data() member, i.e., you also need to take the address of the first element.
0 instead of nullptr. Of course, you need to include the relevant standard headers, too (in this case <algorithm>, <iterator>, <string>, and <vector>).-std=c++11 or -std=c++0x. In this case you might need to define a begin() and end() function.The following worked for me:
const char *argv[]{"-l", nullptr};
execvp("ls", const_cast<char**>(argv));
Since they are string literals, you don't need to null-terminate them, but you still need the null string at the end of the array.
string myargs[5];and you want to pass that toexecve?char *argv[]is actually an array ofchar*elements. Each element in the array points to achar*string. Thereforeargv[0]is achar*string that may be printed/used.