Compilation error while passing double pointer in cpp - c++

Question

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
void f(char **x)
{
(*x)++;
**x = 'a';
}
int main()
{
char str[]="hello";
f(&str);
cout << str << endl;
return 0;
}
Please tell me why this program is giving compilation Error.I am using the g++ compiler
Error :temp1.cpp:16:8: error: cannot convert ‘char (*)[6]’ to ‘char**’ for
argument ‘1’ to ‘void f(char**)’

Answer 1

Arrays can be implicitly converted to pointers, but that doesn't mean that the implicit "pointer equivalent" already exists.
You are hoping that f(&str); will implicitly create both a pointer to str and a pointer to that pointer.
This small (working) change illustrates this point:
int main()
{
char str[]="hello";
char *pstr = str; // Now the pointer extists...
f(&pstr); // ...and can have an address
cout << str << endl;
return 0;
}

Answer 2

You are passing pointer of constant char to the function but in function you are taking it as pointer of pointers. That is the problem. I commented out below where the problem lies.
[Off topic but N. B. : Arrays and pointers are different concept.]
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
void f(char **x) //**x is pointer of pointer
{
(*x)++;
**x = 'a';
}
int main()
{
char str[]="hello";
f(&str); //You are passing pointer of constant char.
cout << str << endl;
return 0;
}

Answer 3

You're going to run into a serious problem with your function f since &str and &str[0] both evaluate to the same value ... as other posters have pointed out, these operations point to different types, but the actual pointer r-value will be the same. Thus in f when you attempt to double-dereference the char** pointer x, you're going to get a segfault even if you attempted something like a cast to massage the type differences and allow compilation to happen with errors. This is because you are never getting a pointer-to-pointer ... the fact that &str and &str[0] evaluate to the same pointer value means that a double-dereference acually attempts to use the char value in str[0] as a pointer value, which won't work.

Answer 4

Your problem is that you're treating arrays as pointers, when they're not. Arrays decay into pointers, and in this case, it doesn't. What you're passing in is a char (*)[6] when it expects a char **. Those are obviously not the same.
Change your parameter to char (*x)[6] (or use a template with a size parameter):
template <std::size_t N>
void f(char (*x)[N])
Once inside, you try to increment what x is pointing to. You can't increment an array, so use an actual pointer instead:
char *p = *x;
p++;
*p = 'a';
All put together, (sample)
template <std::size_t N>
void f(char(*x)[N])
{
if (N < 2) //so we don't run out of bounds
return;
char *p = *x;
p++;
*p = 'a';
}