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:

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

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

        return(0);
}

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

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

Quote:

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Originally Posted by XSquared

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

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*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 :)

Well 4 on common 32-bit desktop machines at least :)

Quote:

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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 :)

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That makes sense. Thanks for the explanation! :)