Convert command line argument to string

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I have a program that reads hard-coded file-path and I want to make it read file-path from command line instead. For that purpose I changed the code like this:

#include <iostream>

int main(char *argv[])
{
...
}

but, argv[1] variable exposed this way seems to be of type pointer, and I need it as a string. What should I do to convert this command line argument to string?

It's already an array of C-style strings:

#include <iostream>
#include <string>
#include <vector>


int main(int argc, char *argv[]) // Don't forget first integral argument 'argc'
{
  std::string current_exec_name = argv[0]; // Name of the current exec program
  std::string first_arge;
  std::vector<std::string> all_args;

  if (argc > 1) {

    first_arge = argv[1];

    all_args.assign(argv + 1, argv + argc);
  }
}

Argument argc is count of arguments plus the current exec file.

c++ - Convert command line argument to string, It's already an array of C-style strings: #include <iostream> #include <string> #​include <vector> int main(int argc, char *argv[]) // Don't forget first integral  I'm working with a command line arguments for the first time and I need to convert one of the string arguments to a character for testing. I tried testing it as a string and it wasn't working. In the program the user needs to enter an e or E or d or D (sample input at top of code).

You can create an std::string

#include <string>
#include <vector>
int main(int argc, char *argv[])
{
  // check if there is more than one argument and use the second one
  //  (the first argument is the executable)
  if (argc > 1)
  {
    std::string arg1(argv[1]);
    // do stuff with arg1
  }

  // Or, copy all arguments into a container of strings
  std::vector<std::string> allArgs(argv, argv + argc);
}

Creating a string from command line arguments, In the last lines of your question, what you mention is C being a procedural language. This means, you should split your semantic logic into reusable smaller​  Parses a Unicode command line string and returns an array of pointers to the command line arguments, along with a count of such arguments, in a way that is similar to the standard C run-time argv and argc values.

No need to upvote this. It would have been cool if Benjamin Lindley made his one-liner comment an answer, but since he hasn't, here goes:

std::vector<std::string> argList(argv, argv + argc);

If you don't want to include argv[0] so you don't need to deal with the executable's location, just increment the pointer by one:

std::vector<std::string> argList(argv + 1, argv + argc);

[PDF] Command-line arguments in the C language, Each argument on the command line is separated by one or more spaces to the character strings containing each argument (char *argv[]). string conversion​. I need to get an argument and convert it to an int. Here is my code so far: Since this answer was somehow accepted and thus will appear at the top, although it's not the best, I've improved it based on the other answers and the comments. The C way; simplest, but will treat any invalid number as 0: The C way with input checking:

It's simple. Just do this:

#include <iostream>
#include <vector>
#include <string.h>

int main(int argc, char *argv[])
{
    std::vector<std::string> argList;
    for(int i=0;i<argc;i++)
        argList.push_back(argv[i]);
    //now you can access argList[n]
}

@Benjamin Lindley You are right. This is not a good solution. Please read the one answered by juanchopanza.

7.13, Command line arguments are optional string arguments that are passed by the up a stringstream variable named convert, initialized with the input from argv[1]. Returns a string array containing the command-line arguments for the current process. An array of string where each element contains a command-line argument. The first element is the executable file name, and the following zero or more elements contain the remaining command-line arguments. The system does not support command-line arguments.

#include <iostream>

std::string commandLineStr= "";
for (int i=1;i<argc;i++) commandLineStr.append(std::string(argv[i]).append(" "));

How to parse command line parameters., argc tells you how many command-line arguments there were. It is always at least 1, because the first string in argv (argv[0]) is the command  As an aside, command-line args are passed as strings anyway. It is odd to literal_eval them like this; a better pattern is to convert the strings to numbers when you need the numbers, and to just leave them as strings otherwise. This will also save you from having to quote-pad the strings twice on the command line.

Command line parameters, of parameters passed to main(), whereas the latter argument, argv, is an array of pointers to character strings which contain these parameters. ftoc.c */ /* Program to convert temperature in Fahrenheit input on command line to temperature  The following example shows how to use command-line arguments in a console application. The application takes one argument at run time, converts the argument to an integer, and calculates the factorial of the number. If no arguments are supplied, the application issues a message that explains the correct usage of the program.

Command-Line Arguments (The Java™ Tutorials > Essential , Parsing Numeric Command-Line Arguments. If an application needs to support a numeric command-line argument, it must convert a String argument that  Command line arguments is a methodology which user will give inputs through the console using commands. Whatever the concept that you preferred to learn in java , we are highly recommended to go through the examples. In reality , theoretical carries a just 20% of the subject , practically carries a lot more than 80%.

Environment.GetCommandLineArgs Method (System), Returns a string array containing the command-line arguments for the current process. Python 3 supports four different ways of handling command line arguments. The oldest one is the sys module. In terms of names, and its usage, it relates directly to the C library ( libc ). The second way is the getopt module which handles both short, and long options, including the evaluation of the parameter values.

Comments
  • The canonical format for main is int main(int argc, char *argv[]). What you have will compile, but it will result in undefined behaviour.
  • I don't need arg count, just the string of representing first argument.
  • hmm.. std::string s(argv[1]); is that what you want to do?
  • @theta: It doesn't matter; it still won't work.
  • "It raises error when I use the program on command line." -- Feel free to be a little more specific.
  • Skip the loop and just use the vector constructor. std::vector<std::string> argList(argv, argv + argc);
  • That's a lot of overhead to extract one string.
  • @BenjaminLindley can you explain what that means? For example, if I didn't want to include argv[0], what would I change in that line? I think this is the kind of initializer being used, but I can't be certain: initializer_list<value_type> il, const allocator_type& alloc = allocator_type() : cplusplus.com/reference/vector/vector/vector
  • Well, my co-worker hooked me up with an answer to my specific request, at least :) std::vector<std::string> args(argv+1, argv + argc); I didn't realize the first parameter was just a pointer, ergo a single increment would be all that's needed.
  • This would create temporary string objects for no reason at all. commandLineStr.append(argv[i]).append(" ") would be much more efficient (append returns *this to allow this chaining). Also, the default constructor will make the string blank, so the initialization is also wasteful.