given a pointer
type * ptr;
When or why would you pass the pointer by refrence to a function ie
function(&ptr);
Instead of simply passing the address directly
function(ptr);
I saw this done and don't understand why.
The prototype for the pass by refrence example
was
function(type **ptr);
A pointer to a pointer. I do not understand.
if the function wanted to change the pointer and not the referenced data it would need to be passed a pointer to the pointer.
Remember that, like any other variable, a pointer itself has an address in memory. When you pass a pointer to a function, a copy of it is made. This copy obviously has a different address than the original. That's why you can do:
The original pointer is unaffected.Code:void print(char * s) { while(*s != 0) { cout << *(s++); } cout << endl; }
Now let's say you wanted to write a function in C that increments a pointer. You would have to do:
With C++, it's simply more readable (and less typing) to use:Code:void inc(char ** p) { *(p++); }
Code:void inc(char * & p) { p++; }
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
I got it so given
type *ptr
ptr is the address of the data the pointer points to or in other words is refrenced by
and
&ptr is the address of the pointer which in turn refrences some data.
