parsing/converting command line arguments into different types

Question

I am trying to write an application that accepts command line argument like 1 + 2, calculates and prints the output. I am very new to c++, so i wrote some code - that's does not work. I think my main problem is lack of knowledge, i cannot figure out how to parse and convert this expression

1 has to go into int/double + has to remain a string to go through the switch statement 2 has to go into int/double

I already know that if there are spaces then each argument will contain the piece of string accordingly having argv[2] = "1" argv[3] = "+" argv = "2" so all i need now is to convert them to the right datatype so i can perform calculations, i have tried to convert using strtod also but every one of them throws exception, plus apparently i cannot write switch statement the way it it so its needs to be a string. I am very confused with all the conversions. Please let me know if i even think in the right way?

 // Calculate.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <string>
#include <tchar.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

int main(int argc, char* argv[])
{
    double oper1 = atof(argv[2]);
    double oper2 = atof (argv[4]);

    switch(char[3]){
    case "+":
        cout<< oper1 + oper2;
        break;
    }


    return 0;
}

There are a number of things wrong with the code. For example, in C/C++, array indexes start from 0, not 1. Double quotes indicate a string, not a character and C++ switch statements do not work on strings. The syntax "char[3]" does not convert the fourth element of the argv array into a char type, either. — akton, Oct 28 '14 at 02:06
@akton At least something that i learned today is that, when you pass command line arguments the actual argument starts at location 2 of an array. It says there is always 1 reserved for the actual command, not sure what 0 is used for. I know the single and double quotes difference, i dont understand why char* acts like string, but if i switch char and change to single quotes it won't work anyway — Jenny, Oct 28 '14 at 02:09
Actually, argv is the parsed command line, so the first element in argv (argv[0]) is usually the executable name. See http://stackoverflow.com/questions/2050961/is-argv0-name-of-executable-an-accepted-standard-or-just-a-common-conventi for more info. — akton, Oct 28 '14 at 02:12
@Jenny The arguments DO start at location 2 in an array. In C++ however this is index 1 (location 1 is index 0 and typically the program name). — Mark B, Oct 28 '14 at 02:14
@MarkB i know this, its not so relevant, i dont get how to do conversions — Jenny, Oct 28 '14 at 02:22

Paul Rooney · Answer 1 · 2014-10-28T02:44:56.730

Try this

// Calculate.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <string>
#include <tchar.h>
#include <stdlib.h>
#include <iostream>
using namespace std;



 int main(int argc, char* argv[])
{
    double coeff1 = atof(argv[1]);
    double coeff2 = atof (argv[3]);
    char op = argv[2][0];

    switch(op){
    case '+':
        cout << coeff1 + coeff2 << endl;
        break;
    default:
        cout << "unhandled operation " << argv[2] << endl;
        break;
    }

    return 0;
}

Note in your code.

switch(char[3]){

Is incorrect, you have no array called char (and you couldn't its a reserved word).

You cannot switch on a string value, so you will have to make it a char.

Your command line arguments are strings. So you read the first character from your argument.

char op = argv[2][0];

and in your switch you can now use

switch(op){

in your case statement

 case "+":

Is wrong. Anything in "" is a string, you must use '' for a char

case '+':

Others have already explained this but your arguments start from argv[1], arg[0] is the name of your program (try printing it to cout). So if you invoke your executable like.

Calculate 1 + 2

The arguments look like

argv[0] = "Calculate"
argv[1] = "1"
argv[2] = "+"
argv[3] = "2"

Note: That this program has zero error checking it will crash if there are not at least 3 command line arguments. You can use argc to determine the number of command line arguments provided.

yeah the error checking totally failed me, my native is java, so i am use to java exception handling...and had no idea its that important to do returns in c++ — Jenny, Oct 28 '14 at 02:48
The return value is not significant here (notice my example always returns 0). You have to ensure you are interpreting the operator (+, - etc) correctly. This was where your initial code was falling down. You can compare chars (my example) or strings (other answer). — Paul Rooney, Oct 28 '14 at 03:00

score 1 · Accepted Answer · answered Oct 28 '14 at 02:36

1

I suggest something like that: (all relevant includes)

using namespace std;

int main(int argc, char* argv[])
{
    if(argc != 4) {
        cout << "usage: calculate <op1> <operation> <op2>" << endl;
        return 1;
    }
    double oper1( atof(argv[1]) );
    double oper2( atof(argv[3]) );
    string operation( argv[2] );
    if(operation == "+") {
        cout << oper1 + oper2;
    } else if (operation == "-") {
        cout << oper1 - oper2;
    } else {
        cout << "illegal operation '" << operation <<"'" << endl;
        return 1;
    }
    return 0;
}

it does not check for a valid floating point number specified

answered Oct 28 '14 at 02:36

mzu

723
3
13

now i see, my total misunderstanding was in exception handling. only because i didn't do returns my program would not compile. You are awesome! – Jenny Oct 28 '14 at 02:44
returns have nothing to do with it. Your program always returned 0. Your program was not compilable, as @Paul Roony pointed out because array with the name of "char" does not exist. "char[3]" does not do what you wanted. You could have used argv[2][0] to extract first character of the 3-rd argument – mzu Oct 28 '14 at 03:19

parsing/converting command line arguments into different types

2 Answers2