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;
}