C++ - char** argv vs. char* argv[]

Solution 1:

They are entirely equivalent. char *argv[] must be read as array of pointers to char and an array argument is demoted to a pointer, so pointer to pointer to char, or char **.

This is the same in C.

Solution 2:

They are indeed exactly the same.

The golden rule of arrays to remember is:

"The name of an array is a pointer to the first element of the array."

So if you declare the following:

char text[] = "A string of characters.";

Then the variable "text" is a pointer to the first char in the array of chars you've just declared. In other words, "text" is of type char *. When you access an element of an array using [index], what you're actually doing is adding an offset of index to the pointer to the first element of the array, and then dereferencing this new pointer. The following two lines will therefore initialize both variables to 't':

char thirdChar = text[3];
char thirdChar2 = *(text+3);

Using the square brackets is a convience provided by the language that makes the code much more readable. But the way this works is very important when you start thinking about more complex things, such as pointers to pointers. char** argv is the same as char* argv[] because in the second case "the name of the array is a pointer to the first element in the array".

From this you should also be able to see why it is that array indexes start from 0. The pointer to the first element is the array's variable name (golden rule again) plus an offset of... nothing!

I've had debates with a friend of mine as to which is better to use here. With the char* argv[] notation it may be clearer to the reader that this is in fact an "array of pointers to characters" as opposed to the char** argv notation which can be read as a "pointer to a pointer to a character". My opinion is that this latter notation doesn't convey as much information to the reader.

It's good to know that they're exactly the same, but for readablity I think that if the intention is an array of pointers then the char* argv[] notation conveys this much more clearly.

Solution 3:

For all practical purposes, they're the same. This is due to C/C++'s handling of arrays passed as arguments, where an array decays to a pointer.