Command-line arguments in C++

[thelimpkid]

I'm trying to use command-line arguments in an efficient way.

When I tried:

int main(int argc, char* argv)
{
	if (argv[1] == "new")
	{
		cout << "Something New!\n";
	}
	return 0;
}

I wasn't able to get "Something New!" to print on the screen.

However, when I tried:

int main(int argc, char* argv)
{
	string cla;
	cla = argv[1];
	if(cla == "new")
	{
		cout << "Something New!\n";
	}
	return 0;
}

I was able to get "Something New!" to print on the screen.

Why didn't it work on the first piece of code?

[drago01]

Because you where comparing a pointer to const char * which could not match, in the second case you are creating a string object and using an overloaded operator for comparison which works exactly this way.

[Jon_Morehouse]

Just to add to drag01's response: In your first case you are asking to compare the string "new" with the contents of argv[1], which is only the character 'n'.

In your second piece of code you are accessing the string capabilities of C++, which know how to deal with an operation (from the comparison operator ==) for a string. Therefore, it reads to the end of the string, stopping when it hits NULL.

[RupertPupkin]

Alternatively, you could do this:

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char* argv[])
{
	if (strcmp(argv[1],"new") == 0)
	{
		cout << "Something New!\n";
	}
	return 0;
}

[thelimpkid]

Thanks for the explanation. It seems so obvious now.
However, I can't get strcmp() to work, even if I include <string>.

Well, it doesn't matter anyway, I want to do it without using a string, and without using a for loop.
What is the best way to capture the command-line arguments?

[drago01]

To use the str* functions include cstring

What is the best way to capture the command-line arguments?

man 3 getopt

[thelimpkid]

getopt() seems like another function I have missed in my book.
Is it just as efficient as using a c-style string?

[TheCodeTwister]

In case of C++, I would prefer:
http://www.boost.org/doc/libs/1_40_0...m_options.html

See the provided example for explanations.

[thelimpkid]

Originally Posted by TheCodeTwister

In case of C++, I would prefer:
http://www.boost.org/doc/libs/1_40_0...m_options.html

See the provided example for explanations.

I think it's quite unnecessary to use an external library for such a simple task. But thanks for pointing into another direction.

I think from what I've seen now is that a c-style string would work in the most efficient way..

Thanks for all the help!

[jonduf]

Originally Posted by thelimpkid

I think it's quite unnecessary to use an external library for such a simple task. But thanks for pointing into another direction.

I think from what I've seen now is that a c-style string would work in the most efficient way..

Thanks for all the help!

In your example maybe, but as the number of options grow and the complexity of the arguments grow, it is best to use a well understand and well tested library that is designed to handle command line args properly. Plus it leaves you the opportunity to code the meat and potatoes of your program.

[thelimpkid]

Originally Posted by jonduf

In your example maybe, but as the number of options grow and the complexity of the arguments grow, it is best to use a well understand and well tested library that is designed to handle command line args properly. Plus it leaves you the opportunity to code the meat and potatoes of your program.

I noticed indeed that it can be handy to use a well tested library when the number of options grow. In terms of performance it doesn't matter at all I guess, so after all it doesn't seem so useless to use an external library at all.

[mohan.10]

Originally Posted by thelimpkid

I'm trying to use command-line arguments in an efficient way.

When I tried:

int main(int argc, char* argv)
{
	if (argv[1] == "new")
	{
		cout << "Something New!\n";
	}
	return 0;
}

I wasn't able to get "Something New!" to print on the screen.

However, when I tried:

int main(int argc, char* argv)
{
	string cla;
	cla = argv[1];
	if(cla == "new")
	{
		cout << "Something New!\n";
	}
	return 0;
}

I was able to get "Something New!" to print on the screen.

Why didn't it work on the first piece of code?

Hey here are the exact lines from the book i'm referring to learn c++..

"Testing how C++ compares quoted strings is more difficult. Instead of using the contents
of the string, the compiler uses the location of the strings in memory, which is a detail of the
compiler’s internal workings and has no bearing on anything practical. Thus, unless you know
how the compiler works, you cannot predict how it will compare two quoted strings. In other
words, don’t do that. Make sure you create
objects before you compare strings."

It seems you are learning c++ too..I can suggest "Exploring C++(Apress)"..