Simple Calculator using command line argument C++ - c++

Question

I'm new to C++. I'm writing a simple calculator using command line. The command line should have this format:
programname firstNumber operator secondNumber
Here what I got so far:
#include <iostream>
#include <fstream>
using namespace std;
int main(int argc, char* argv[])
{
if (argc != 3)
{
cerr << "Usage: " << argv[0] << endl;
exit(0);
}
else
{
int firstNumber = atoi(argv[1]);
char theOperator = atoi(argv[2]);
int secondNumber = atoi(argv[3]);
switch (theOperator)
{
case'+':
{
cout << "The answer is " << firstNumber + secondNumber << endl;
break;
}
case '-':
{
cout << "The answer is " << firstNumber - secondNumber << endl;
break;
}
case '*':
{
cout << "The answer is " << firstNumber * secondNumber << endl;
break;
}
case '/':
{
if (secondNumber == 0)
{
cout << "Can not devide by a ZERO" << endl;
break;
}
else
{
cout << "The answer is " << firstNumber / secondNumber << endl;
break;
}
}
}
}
}
The program does not run. When I run it, it displays an appropriate usage message and end the program. Can anyone please help me?

Answer 1

Others have already given you the answer but you could have very easily figured this one out on your own. Just print what argc is at the point where you know the code is going into:
int main(int argc, char* argv[])
{
if (argc != 3)
{
cout << "argc is: " << argc << endl; // Debug output that you delete later
cerr << "Usage: " << argv[0] << endl;
exit(0);
}
else
And then come back with what argc is. When you find that argc is actually 4 and you want to know what is inside argc you should write some code to print it so that you can figure it out... Like this:
int main(int argc, char* argv[])
{
cout << "argc is: " << argc << endl; // Debug output that you delete later
for (int i = 0; i < argc; ++i)
{
// Print out all of the arguments since it's not working as you expect...
cout << "argv[" << i << "] = " << argv[i] << endl;
}
if (argc != 3)
{
cerr << "Usage: " << argv[0] << endl;
exit(0);
}
else
and you would have very quickly figured out what is wrong...
Please learn how to do this because it will save your but in the future and you won't have to wait for an answer here.
Additionally there is another error in your code.
Why on earth are you converting the + character from a string to an int?
else
{
int firstNumber = atoi(argv[1]);
char theOperator = atoi(argv[2]); // <<< WTF? Why?
int secondNumber = atoi(argv[3]);
switch (theOperator)
You probably want to get rid of the atoi part there and just go with:
char theOperator = argv[2][0]; // First character of the string
Provided that the second argument will always have only one letter... Which you might want to enforce/check. See strlen() and std::string and note that the type of argv[2] is char* (pointer to char).
I also recommend that you read How to debug small programs which is linked from the SO Howto-Ask Help Page. It may help a little. And no, I don't think your question is bad. Debugging small programs is a skill you'll need in the future if you intend to program so it will benefit you to learn it now.
Welcome to programming and C++ :)

Answer 2

The parameter argc also counts the program's name.
Try this:
if (argc != 4) // We expect 4 arguments: programname number operator number
{
cerr << "Usage: " << argv[0] << " <number> <operator> <number>" << endl;
exit(0);
}
In your code, running the program correctly (with all 3 parameters) displays the error message because argc equals 4.

Answer 3

If you type in
programname firstNumber operator secondNumber
You have 4 arguments, not 3.
argv[0] = programname
argv[1] = firstNumber
argv[2] = operator
argv[3] = secondNumber
Looks like your program is working correctly... at least as far as printing the usage message goes.
See also the other comments regarding your use of the operator argument.