What are the distinctions between the various symbols (*,&, etc) combined with parameters? [duplicate]

Answer 1

The symbols & and * are used to denote a reference and pointer type, respectively.

int means simply the type 'int', int* means 'pointer to int', int& means 'reference to int',

A pointer is a variable which is used to store the address of a variable. A reference has the syntax of its base type, but the semantics of a pointer to that type. This means you don't need to dereference it in order to change the value.

To take an example, the following code blocks two are semantically equivalent:

int* a = &value;
*a = 0;

And:

int& a = value;
a = 0;

The main reasons to use pointers or references as an argument type is to avoid copying of objects and to be able to change the value of a passed argument. Both of these work because, when you pass by reference, only the address is copied, giving you access to the same memory location as was "passed" to the function.

In contrast, if a reference or pointer type is not used, a full copy of the argument will be made, and it is this copy which is available inside the function.

Answer 2

To understand this you'll first need to understand pointers and references. I'll simply explain the type declaration syntax you're asking about assuming you already know what pointers and references are.

In C, it is said that 'declaration follows use.' That means the syntax for declaring a variable mimics using the variable: generally in a declaration you'll have a base type like int or float followed something that looks like an expression. For example in int *y the base type is int and the expression look-alike is *y. Thereafter that expression evaluates to a value with the given base type.

So int *y means that later an expression *y is an int. That implies that y must be a pointer to an int. The same holds true for function parameters, and in fact for whole function declarations:

int *foo(int **bar);

In the above int **bar says **bar is an int, implying *bar is a pointer to an int, and bar is a pointer to a pointer to an int. It also declares that *foo(arg) will be an int (given arg of the appropriate type), implying that foo(arg) results in a pointer to an int.¹ So the whole function declaration reads "foo is a function taking a pointer to a pointer to an int, and returning a pointer to an int."

C++ adds the concept of references, and messes C style declarations up a little bit in the process. Because taking the address of a variable using the address-of operator & must result in a pointer, C doesn't have any use for & in declarations; int &x would mean &x is an int, implying that x is some type where taking the address of that type results in an int.² So because this syntax is unused, C++ appropriates it for a completely different purpose.

In C++ int &x means that x is a reference to an int. Using the variable does not involve any operator to 'dereference' the reference, so it doesn't matter that the reference declarator symbol clashes with the address-of operator. The same symbol means completely different things in the two contexts, and there is never a need to use one meaning in the context where the other is allowed.

So char &foo(int &a) declares a function taking a reference to an int and returning a reference to a char. func(&x) is an expression taking the address of x and passing it to func.

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_{1. In fact in the original C syntax for declaring functions 'declarations follow use' was even more strictly followed. For example you'd declare a function as int foo(a,b) and the types of parameters were declared elsewhere, so that the declaration would look exactly like a use, without the extra typenames.}

_{2. Of course int *&x; could make sense in that *&x could be an int, but C doesn't actually do that.}

Answer 3

Ex: What is the difference between 'void func(int *var)' and 'void func(int **var)'? What about 'int &var'?

The same question stands for return types, as well as arguments. What does 'int& func(int var)' mean, as compared to 'int* func(int var)'? And in arguments, how does 'y = func(*x)' differ from 'y = func(&x)'?

(1)
    <return type> <function name> <parameters> 
    void           func            (int *var) 

    <parameter> here int *var is a pointer to integer, ie it can point to 
    an array or any buffer that should be handled with integer pointer 
    arithmetic. In simple terms , var holds the address of the respective 
    **actual parameter.**

    eg: int arr[10]; 
        func(arr);

        int a = 33;
        func(&a);

        Here, &a means we are explicitly passing address of the the variable 'a'. 
 (2) 
     int m = 0; 
     int &var = m;

     Here var means reference, ie it another alias name for variable 'm' , 
     so any change 'var' makes will change the contents of variable 'm'.

     var = 2; /* will change the actual contents of 'm' */ 

     This simple example will not make sense , unless you understand the context. 
     Reference are usually use to pass parameter to function, so that changes made by 
     the function to the passed variable is visible at the caller.

     int swap(int &m, int &n) {
         tmp = m;
         m = n; 
         n = tmp; 
     }   

     void main( void ) {

        int x = 1, y = 2;
        swap(x, y);
        /* x = 2, y =1 */   

     }

(3) 
     'int& func(int var)' mean, as compared to 'int* func(int var)'?

      int& func(int var) means the function returns a reference; 
      int* func(int var) means the function returns a pointer / address; 

      Both of the them has its context; 


      int& setNext() {

            return next;   
      } 

      setNext() = previous; 


      where as 

      int* setNext() {

           return &next; 
      } 

      int *inptr;
      inptr = setNext(); 
      *inptr = previous;

      In the previous two lines, 

      int *inptr <- integer pointer declaration; 
      *inptr <- means we are accessing contents of the address pointed by inptr; 
      ie we are actually referring to 'next' variable. 

      The actual use is context specific. It can't be generalized.  

(4) 
    how does 'y = func(*x)' differ from 'y = func(&x)'?                           

     y = func(&x) is already explained. 

     y = func(*x) , well i'm not sure if you actually meant *x.


     int swap(int *m, int *n) {
         tmp = *m;
         *m = *n; 
         *n = tmp; 
     }   

Answer 4

The symbols * and & have three meanings each in C++:

1. When applied to an expression, they mean "dereference" and "address-of" respectively, as you know.

2. When part of a type, they mean "pointer" and "reference", respectively.

   Since C++ doesn't care about arbitrary spacing, the declaration int *ptr is exactly the same as the declaration int* ptr, in which you can now more clearly see that this is an object called ptr of type int*.¹

3. When used between two expressions, they mean "multiply" and "bitwise AND", respectively.

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^{1 - though, frustratingly, this isn't actually how the internal grammar reads it, thanks to the nasty legacy of C's type system. So avoid single-line multi-declarations involving pointers unless you want a surprise.}

Answer 5

What you're asking about are called pointers (*), and reference to (&), which I think is best explained here.