Passing a char pointer array to a function

Question

I have written following sample code to demonstrate my problem

#include 
#include 

using namespace std;

void f (char*** a)
{
           


        

Answer 1

It's because of the operator precedence, where the array-indexing operator [] have higher precedence than the dereference operator *.

So the expression *a[0] is really, from the compilers point of view, the same as *(a[0]), which is not what you want.

You have to explicitly add parentheses to change the precedence:

(*a)[0] = ...

Answer 2

a[0], a[1] are char, and they have a value in them, dereferencing that value will obviously cause a segmentation fault. What you may want to do is:

(*a)[...]

dereference 'a' which is a pointer, this will give u an array.