Is Str a "pointer" or a "pointer to pointer" in Str[3][5]?
I usually declare char **Str to make a simulated 2D array, but how to send the address of Str[3][5] to another function, and how to set the parameter of that function in order to modify values in Str[3][5] ?
As far as I know, the compiler treats a 2D array as an ordinary array, so I should send Str (a pointer) to the other function. But how to modify the value of Str[3][5] via the received pointer?
An array of T when used in most expressions will decay to a pointer of type "pointer to T" and have the value equal to the address of the first element of the array.
If Str is declared as:
char Str[3][5];
Then Str will decay to char (*)[5], and have the value of &Str[0]. If passing to a function, you can declare the function as follows:
void foo (char (*str)[5], size_t n);
And within foo(), you would access str as you would access Str itself.
Pointer address in a C multidimensional array
To expand on this answer to a similar question, when we type to get addresses of arrays. The name of the array is actually a pointer to the first variable in the array.
You can't really pass a pointer of a particular part of an array, as what we interact with in coding is the pointer itself. (array + 1) == array[1] Let's say we have create an array pointer int* arrayPointer; and you want to get the second variable in the array. You could either assign arrayPointer = array + 1 or arrayPointer = &array[1]
So if you want to modify part of array[3][5] the easiest way is to pick that part of the array and assign a variable to it. For example if you want to modify the first "column" and first "row" you would type...
array[0][0] = variable
Hope this helps.
