about strings in c++

Question

Hey, guys
Can anyone help me? I can't understand why in this code i have to use &str[2]. How it works? Why if i remove & the compiler gives an error converting char to const char *? and why everything is ok with the &?

#include <iostream>
#include <string.h>
#define size 255
using namespace std;
int main()
{
char *str, *newstr;
str= new char[size];
newstr= new char[size];
cout<<"vvedite stroku:"<<endl;
cin.getline(str,size);
strncpy(newstr,&str[2],5);
}

Answer 1

strmcy, copies a character using a pointer, hence `const char*`

When you pass a character into every function, the char will be copied, thus taking some memory.

When a function receives an address (by using &), the char will not be copied. Hence, there will be less memory usage.

Consider the following:

void add(int a, int b) {
// a and b are copies of the original variable

cout << a + b;
}

void add2(const int* a, const int* b) {
// a and b have the same address as the original variables
// The original variables are therefore not copied

cout << *a + *b;
}

int main() {
int v1 = 1;
int v2 = 1;

add(v1, v2);
add2(&v1, &v2);

return 0;
}

----

Which one is more memory efficient? add2 because the local variables, a and b, have the same address as v1 and v2 respectively.

Answer 2

Apart from memory efficiency, when you pass `str[2]` you are passing a char rather than a char pointer (which is required by `strncpy` function like Edwin said). Because the function needs a pointer to work; you should pass the address of the char, not the char itself. And that's why you add & operator before `str[2]`, so the function gets the address rather than the char. `&str[2]` means the address of pointer `str` + 2 * sizeof(char).

You can still pass the pointer simply by adding a value to the pointer, like this:

`strncpy(newstr, str + 2, 5);`

And here you don't need the & operator anymore, because this way you are passing the pointer.

Hth, cmiiw