command line help

[pilotnate]

Hey everyone, I am trying to use command line arguments. The program should take 5 integers via the command prompt and then add all the numbers and print the sum. It is not working. I got a previous version to compile and run, but It was not correct. It essentially ran the program from the command propmt, instead of taking the arguments autmatically

  #include <iostream>
#include <limits>

using namespace std;

int main(int argc, char *argv[])
{
    //if the user did not enter enough arguments
    if(argc < 2)
    {
        //type out what is required
        cerr << "Usage: " << argv[0] << " integers" << endl;
        return 1;
    }

    int input[5];       //array declaration
    for(int i = 0; i < argc; ++i)
    {
        cout << i + 1 << ").  ";
        //input validation
        if((double)argv[i] != (int)arv[i])
        {
            cerr << "ERROR: Arguments not valid\n";
            return 1;
        }
        input[i] = argv[i];     //initializes each array index
    }

    int sum = 0;        //sum of the values entered
    //sums the integers entered
    for(int i = 0; i < 5; ++i)
        sum += input[i];

    cout << "Sum of all integers entered = " << sum << endl;

    return 0;
}

[kevinkjt2000]

Use the atoi function to get integers and atof to get floating point numbers from argv. See an example here: http://www.cs.ucsb.edu/~pconrad/cs16/topics/cmdlineargs/
Casting an array of chars to int or to double is not supported in c++.

[admkrk]

The program should take 5 integers

You have:

    //if the user did not enter enough arguments
    if(argc < 2)
    {
        //type out what is required
        cerr << "Usage: " << argv[0] << " integers" << endl;
        return 1;
    }

Besides your Usage: output being wrong, if it requires 5 integers and the user input 4 the error would not be displayed with if(argc < 2). They only need to enter two integers to pass that test. Usage: only asks for one integer.

[chipp]

what this function used for?

if((double)argv[i] != (int)arv[i])

if you want to convert char * into int you should just convert it and not compare it

CMIIW

[kevinkjt2000]

@chipp: I believe the OP is trying to see if the number is a whole number.

[dhayden]

If you really want to make sure they've entered an integer, use strtol()

char *endp;
long val;
...
val = strtol(argv[i], &endp, 10);
if (*endp) {
    // user didn't enter a number
}

This will work properly if they enter odd things like "-1" or "-123hello" or "" or "18.0"

[kevinkjt2000]

I figured why not test the two functions?
strtol_vs_atoi.cpp:

#include <iostream>
#include <iomanip>
#include <cstdlib>
using namespace std;

int main(int argc, char* argv[]) {
	
	for(int i = 1; i < argc; i++) {
		cout << "argv[" << i << "]: " << argv[i] << "\n";
		cout << "atof: " << atof(argv[i]) << "\n";
		cout << "atoi: " << atoi(argv[i]) << "\n";
		cout << "integer? " << boolalpha << (atof(argv[i]) == atoi(argv[i])) << "\n";
		cout << "strtol: " << strtol(argv[i], NULL, 0) << "\n\n";
	}
	
	return 0;
}

How I compiled and ran the program:
g++ strtol_vs_atoi.cpp -o a.out && ./a.out -1 -123hello 18.0 19.6

The output I received:

argv[1]: -1
atof: -1
atoi: -1
integer? true
strtol: -1

argv[2]: -123hello
atof: -123
atoi: -123
integer? true
strtol: -123

argv[3]: 18.0
atof: 18
atoi: 18
integer? true
strtol: 18

argv[4]: 19.6
atof: 19.6
atoi: 19
integer? false
strtol: 19

In summary:
Teaching atoi to beginners is just easier. There will be no chance of them being confused by extra parameters. I say this even though I use strtol for better error checking in my production code.

Edit: I changed the code to also show detection of whether or not the number is an integer.

[pilotnate]

Wow guys thanks! I'll definitely do some research about this today and figure it out. I appreciate all the feedback! I'll update when I finish

[pilotnate]

Alright this works great!!

#include <iostream>
#include <cstdlib>
#include <limits>

using namespace std;

int main(int argc, char *argv[])
{
    //if the user did not enter enough arguments
    if(argc != 6)
    {
        //type out what is required
        cerr << "Usage: " << argv[0] << " int1 int2 int3 int4 int5\n";
        return 1;
    }

    int input[5];       //array declaration
    for(int i = 1; i < argc; ++i)
    {
        input[i-1] = atoi(argv[i]);     //converts input to int
    }

    int sum = 0;        //sum of the values entered
    //sums the integers entered
    for(int i = 0; i < 5; ++i)
        sum += input[i];

    cout << "Sum of all integers entered = " << sum << endl;

    return 0;
}

Again thanks for all the help!!