Website tutorial; Command line arguments

[TheUnknownFacto]

SO yeah I'm still working on the tutorial, still happy with it in general Now I came to lesson 14 about command line arguments. It's all pretty simple, and I didn't have much of any problem going through the lesson, and writing a small application derived of the code example given.

However hwn I quized myself I messed up on one of the questions.

...

int main ( int argc, char *argv[] )

...

3. What type is argv?
A. char *
B. int
C. char **
D. It's not a variable

Now the quiz tells me that the type of argv is "char**", could someone explain to me the difference between "char *" and "char **", and if not entirely obvious, why "char **" is the correct anwser.

Very much appreciated.

[laserlight]

A char* is a pointer to a char. A char** is a pointer to a pointer to a char.

Now, arrays decay to a pointer to their first element when passed as arguments to functions. As such, the "char* argv[]" parameter of main() actually ends up as a char**, though we think of it as an array of null-terminated strings of characters.

[indigo0086]

char* argv[] is the same as char** argv.

char* argv[] indicates it's an array of pointer to char. Strings are pretty much a null terminated set of sequential characters, with char* indicating a pointer to the address of the first character. Pointers in themselves can be used to reference arrays using pointer arithmetic.

char** argv indicates it's a pointer to char*'s. So it's the first address of an array of pointers to characters.

[TheUnknownFacto]

I think I'm starting to understand it (though it's still quite confusing in my mind).

char* indicates a pointer to a something.
argv[] indicates an array

So, the type of argv[] would be char*, however the type of argv's first element would be char**.

If I'm wrong to this point, it more or less makes my next question stupid, if I'm right to this point then...

I'm still a bit confused about how "argv" evaluates to the first item within the argv[] array.

[laserlight]

So, the type of argv[] would be char*, however the type of argv's first element would be char**.

I think you got it the other way round. By virtue of being an array of strings, argv is a char**. The first element of argv is a string, and thus is a char*.

I'm still a bit confused about how "argv" evaluates to the first item within the argv[] array.

Consider this example program:

#include <iostream>

void tellSize(int nums[])
{
    std::cout << sizeof(nums) << std::endl;
}

int main()
{
    int numbers[20];
    std::cout << sizeof(numbers) << std::endl;
    tellSize(numbers);
}

Assuming that you are working with 4 byte ints (and pointers), you will find that the output of that program is:
80
4

In the main() function, numbers is an array of 20 ints, and thus its size in bytes in 20 * 4 = 80. When it is passed to tellSize(), the nums is actually a pointer to the first element of numbers. Since it is a pointer, its size is 4 bytes, not 80 bytes, although the array it points to is still 80 bytes in size.

[indigo0086]

to illustrate the address location

#include <iostream>

using namespace std;

void printFirst(int* first)
{
    cout << *first << endl;
}

int main()
{
	int arr[4] = {1, 2, 3, 4};
	printFirst(arr);
	return 0;
}

printFirst takes a pointer to int. In main I create an array with 4 elements. Then when I call printFirst I pass it the variable name, which is actually the address of the first element in the array. I would get the same effect as calling printFirst(&arr[0]);

if I wanted to iterate through the array I owuld just increment the pointer in the printFirst Method as so

void printFirst(int* first)
{
    for(int i = 0; i < 4; ++i)
    {
        cout << *first << " ";
        ++first;
    }
    cout << endl;
}