Getting the size of **argv.

This is a discussion on Getting the size of **argv. within the C++ Programming forums, part of the General Programming Boards category; No matter what I try to do here, I always come out with a result of 4 when I do ...

  1. #1
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    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.

  2. #2
    C++ Developer XSquared's Avatar
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    #include <cstring> and use strlen( argv[1] ).
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  3. #3
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    Quote Originally Posted by XSquared
    #include <cstring> and use strlen( argv[1] ).
    *looks dumbfounded*

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

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    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

  5. #5
    and the hat of wrongness Salem's Avatar
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    Well 4 on common 32-bit desktop machines at least
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    Quote 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!

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