C++: how to read **argv value and copy into a string - c++

Question

I'm making a new app for my lab and I must take form terminal some arguments, and I want to copy the **args value into a string, and after that I must match the args for execute some foreplay operation.
I try to build this code to copy the **argv value into string var but this is the error:
No source available for "std::string::operator=() at 0x44c1dc"
The code is this:
int main(int argc, char **argv)
{
string args[argc-1];
int j=0;
for(int i=2; i<argc-1;i++)
{
j=0;
while(argv[i][j]!='\0')
{
args[i]=args[i]+argv[i][j];
j++;
}
}
return 0;
}

Answer 1

I would propose this solution:
std::vector<std::string> args;
std::copy(argv + 1, argv + argc, std::back_inserter(args));
References:
std::copy
std::back_inserter
To explain it a little, this copies from argv[1] (I don't care about argv[0] as it's the "command" and usually irrelevant for argument handling) into the vector. We stop copying at argv[argc] which is one beyond the end of the actual arguments (which is how "end" iterators work).
std::back_inserter creates a special iterator that is used to insert ad the end of containers.

Answer 2

Fastest and shortest solution with range constructor:
std::vector<std::string> args(argv + 1,argv + argc);

Answer 3

Make things as clear as you can. (Will help those 3am debugging sessions in your laboratory).
int main(int argc, char **argv)
{
std::vector<std::string> args(argc);
for (int i = 0; i < argc; ++i){
args[i] = argv[i];
}
// rest of code here.
}
Yes I know I waste args[0], but do you really care about that? The indexing is so much clearer if you do it this way, args has the same indexing conventions as argv (which all us old cats are used to), and you don't get a whole lot of edge cases when you only pass the program name.

Answer 4

There are a few mistakes in your code, which also causes the odd problems. Also you're overcomplicating things quite a bit.
Here's what I'd do, this will require one additional header, <vector>, for the STL container, but it will make things a lot easier (since you shouldn't rely on being able to create arrays with dynamic lengths (also note that your array length might be 0, if there are no arguments given!))
int main(int argc, char **argv) {
std::vector<const std::string>> params; // This is the container that will take all arguments.
// Iterate over all arguments provided
// (skipping the very first, since that's the executable itself)
for (int i = 1; i < argc; i++)
// Add the argument to the vector
// (this will construct a new std::string and add it to the end).
params.push_back(argv[i]);
// You're now able to access the arguments passed:
// params.size() will return the number of arguments.
// params[i] will return the i-th argument, starting at 0.
return 0;
}

Answer 5

int main(int argc, char **argv)
{
string args[argc];
int j=0;
for(int i=2; i<argc;i++)
{
args[i-2].assign(argv[i]);
}
return 0;
}

Answer 6

first, you can't use a dynamic argument in a static expression.
you must use either new to create a string array, or, better yet, use a vector (or any other container you like).
second, there is no point in copying the values one character by another. copy the entire string.
here is a sample code that will work:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
typedef vector<string> svec;
int main(int argc,char **argv)
{
svec strings;
if(argc>1)
{
cout <<"arguments:"<<endl;
for(int i=1;i<argc;i++)
{
strings.push_back(argv[i]);
}
}
for(auto i:strings)
{
cout << i;
}
return 0;
}
D.