编写一个程序，该程序读取所有命令行参数并打印平均值，c ++

Question

//review3
#include <iostream>
#include <cstdlib>
#include <string>
#include <fstream>
using namespace std;

double average;
int main(int argc, char* argv[])
{
    int num1 = atoi (argv[1]);
    int num2 = atoi (argv[2]);
    int num3 = atoi (argv[3]);
    average = ((num1 + num2 + num3) / 3);
    cout << average << endl;
}

I am not sure how to tackle this problem if I need to calculate the average of all of the command line arguments? This is how I would do it with 3 CLA's but I am unsure how I would do it without knowing a set amount of CLA's for this problem. Does anyone also know how to do this if you had to find the median?

Answer 1

Below is the simple program.

int sum = 0 ;
for ( int i = 1; i < argc; i++ )
{
  sum = sum + atoi(argv[i]) ;  //Exclusion of argv[0] is no incidence...
}

Then you can do whatever you want to do with sum

Answer 2

In the declaration of main int argc is the number of arguments given. You can use a for loop to iterate through your arguments and calculate the average:

// Your includes here
// ...

int main(int argc, char* argv[])
{
    int average = 0;

    for(int i = 1; i < argc; i++) // argv[0] is the name of your program, so we are skipping it
    {
        average += atoi(argv[i]);
    }

    average = average / (argc - 1);
}

Answer 3

argc will give you the number of arguments given to main().

Note that argc will be ONE, not zero if you provide no argument as argv[0] is the file name of the executable.

To get the median, the following should do the trick:

float median;

// Get a sorted list of the integers
std:list<int> args;
for(int i = 1; i < argc; i++)
{
    args.push_back(atoi(argv[i]));
}
args.sort();

// Extract median from the sorted array of integers
int middle_index = args.size() / 2;
if (args.size() % 2 == 1)
{
    // For odd number of values, median is middle value
    median = args[middle_index];
}
else
{
    // For even number of values, median is average of the two middle values
    median = (args[middle_index-1] + args[middle_index]) / 2.0;
}