Thread: Getting the size of **argv.

No matter what I try to do here, I always come out with a result of 4 when I do a sizeof(argv[1]). No matter how large or small the input is on the command line, the sizeof always returns 4.

My question now is, how can I get the size of the array that holds the char data from the input of the command line?

Some code I tried:

Code:

int main(int argc, char **argv)
{
	int x = sizeof(argv[1]);
	std::cout << x << "\n";

	x = sizeof(argv);
	std::cout << x << "\n";

	return(0);
}

And the output? 4 on both lines. It's starting to confuse me.

Originally Posted by XSquared

#include <cstring> and use strlen( argv[1] ).

*looks dumbfounded*

How'd I forget that? Heh.. Thank you!

The reason you were getting 4 as a result, was that you were essentially performing sizeof(char*), which of course will be 4 on a 32 bit machine

Originally Posted by bithub

The reason you were getting 4 as a result, was that you were essentially performing sizeof(char*), which of course will be 4 on a 32 bit machine

That makes sense. Thanks for the explanation!