C ++ - 接受命令行参数？

Question

So this was my original code:

#include <iostream>

using namespace std;

int main ()
{
    float x;  
    cout << "Please enter an integer value: ";
    cin >> x;

    if ((x >= 100) && (x < 200)) {
        cout << "split";
    } else if (x == 0 ||x == 1 ) {
        cout << "steal";
    } else {
        cout << "split";
    } 

    system("pause");
}

---

It works perfectly, but I need it to run this way:

C:\> program.exe 109

it will read 109 and give the output - "steal" .

C:\> program.exe 0.5

it will read 0.5 and give me the output "split" .

What do I have to add to my original code to do this?

Answer 1

Change your main to

int main (int argc, char** argv)

The you can check number of specified parameters to your program in argc and the values (as char * ) in argv . You can convert that values to float using std::stof

float x = 0.0f;
if (argc > 1) {
    x = std::stof(argv[1]);
} else {
    std::cerr << "Not enough arguments\n";
    return 1;
}

Please note that the first argument to the program is the name of the executable itself ( program.exe in your case), so you need to check for at least two arguments.

References: http://en.cppreference.com/w/cpp/string/basic_string/stof

Answer 2

you can do this by using command line arguments. Here is format for main function:

int main (int argc, _TCHAR* argv[])
{
}

Here argc represents number of arguments(returns 2 in your case, program.exe 0.5 )

argv represents two strings. 1st containing program.exe and second containing 0.5 .

This way you can solve you problem

Answer 3

Can we have more clarity on the question? Do you want to know how to execute the code using command line arguments? In that case here it is:

int main (int no_of_args, char* arglist[])
{
}

In argList , the first item holds the name of the executable and subsequent items hold the inputs provided.