Checking for specific command line arguments in C++

Question

I am building a CLI app that should do something similar to this:

./app

Welcome to the app, Type -h or --help to learn more.

./app -h

list of commands:...

Here is the code i am trying to build:

#include <iostream>

using namespace std;

int main(int argc, char** argv)  {

   cout << "Welcome to the app. Type -h or --help to learn more\n";

   if(argv == "-h" || argv == "--help") {
      cout << "List of commands:...";
   }
  return 0;
}

But when i try to compile gcc gives following erros:

error: comparison between distinct pointer types ‘char**’ and ‘const char*’ lacks a cast [-fpermissive]
    if(argv == "-h" || argv == "--help") {
               ^~~~
error: comparison between distinct pointer types ‘char**’ and ‘const char*’ lacks a cast [-fpermissive]
    if(argv == "-h" || argv == "--help") {
                               ^~~~~~~~

Answer 1

从C ++ 17开始，编写此代码的最佳方法如下：

#include <iostream>
#include <string_view>

int main(int argc, char** argv) {
    using namespace std::literals;

    std::cout << "Welcome to the app. Type -h or --help to learn more\n";

    if (argv[0] == "-h"sv || argv[0] == "--help"sv) {
       std::cout << "List of commands:...";
    }
}

在string_view标头存在之前，您可以使用""s std::string文字，它会产生与上面相同的代码，只包括string标准标头并更改"…"sv为"…"s。不幸的是，这样的代码会导致多余的分配，但是在这个特定示例中这是不相关的。