What's the difference between * and & in C?

Question

I'm learning C and I'm still not sure if I understood the difference between & and * yet.

Allow me to try to explain it:

int a; // Declares a variable
int *b; // Declares a pointer
int &c; // Not possible

a = 10;
b = &a; // b gets the address of a
*b = 20; // a now has the value 20

I got these, but then it becomes confusing.

void funct(int a) // A declaration of a function, a is declared
void funct(int *a) // a is declared as a pointer
void funct(int &a) // a now receives only pointers (address)

funct(a) // Creates a copy of a
funct(*a) // Uses a pointer, can create a pointer of a pointer in some cases
funct(&a) // Sends an address of a pointer

So, both funct(*a) and funct(&a) are correct, right? What's the difference?

Answer 1

* and & as type modifiers

• int i declares an int.
• int* p declares a pointer to an int.
• int& r = i declares a reference to an int, and initializes it to refer to i.
  _{C++ only. Note that references must be assigned at initialization, therefore int& r; is not possible.}

Similarly:

• void foo(int i) declares a function taking an int (by value, i.e. as a copy).
• void foo(int* p) declares a function taking a pointer to an int.
• void foo(int& r) declares a function taking an int by reference. (C++ only)

* and & as operators

• foo(i) calls foo(int). The parameter is passed as a copy.
• foo(*p) dereferences the int pointer p and calls foo(int) with the int pointed to by p.
• foo(&i) takes the address of the int i and calls foo(int*) with that address.

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(tl;dr) So in conclusion, depending on the context:

• * can be either the dereference operator or part of the pointer declaration syntax.

• & can be either the address-of operator or (in C++) part of the reference declaration syntax.

  _{Note that * may also be the multiplication operator, and & may also be the bitwise AND operator.}