Why use double indirection? or Why use pointers to pointers?

Question

When should a double indirection be used in C? Can anyone explain with a example?

What I know is that a double indirection is a pointer to a pointer. Why would I ne

Answer 1

The following is a very simple C++ example that shows that if you want to use a function to set a pointer to point to an object, you need a pointer to a pointer. Otherwise, the pointer will keep reverting to null.

(A C++ answer, but I believe it's the same in C.)

(Also, for reference: Google("pass by value c++") = "By default, arguments in C++ are passed by value. When an argument is passed by value, the argument's value is copied into the function's parameter.")

So we want to set the pointer b equal to the string a.

#include 
#include 

void Function_1(std::string* a, std::string* b) {
  b = a;
  std::cout << (b == nullptr);  // False
}

void Function_2(std::string* a, std::string** b) {
  *b = a;
  std::cout << (b == nullptr);  // False
}

int main() {
  std::string a("Hello!");
  std::string* b(nullptr);
  std::cout << (b == nullptr);  // True

  Function_1(&a, b);
  std::cout << (b == nullptr);  // True

  Function_2(&a, &b);
  std::cout << (b == nullptr);  // False
}

// Output: 10100

What happens at the line Function_1(&a, b);?

• The "value" of &main::a (an address) is copied into the parameter std::string* Function_1::a. Therefore Function_1::a is a pointer to (i.e. the memory address of) the string main::a.

• The "value" of main::b (an address in memory) is copied into the parameter std::string* Function_1::b. Therefore there are now 2 of these addresses in memory, both null pointers. At the line b = a;, the local variable Function_1::b is then changed to equal Function_1::a (= &main::a), but the variable main::b is unchanged. After the call to Function_1, main::b is still a null pointer.

What happens at the line Function_2(&a, &b);?

• The treatment of the a variable is the same: within the function, Function_2::a is the address of the string main::a.

• But the variable b is now being passed as a pointer to a pointer. The "value" of &main::b (the address of the pointer main::b) is copied into std::string** Function_2::b. Therefore within Function_2, dereferencing this as *Function_2::b will access and modify main::b . So the line *b = a; is actually setting main::b (an address) equal to Function_2::a (= address of main::a) which is what we want.

If you want to use a function to modify a thing, be it an object or an address (pointer), you have to pass in a pointer to that thing. The thing that you actually pass in cannot be modified (in the calling scope) because a local copy is made.

(An exception is if the parameter is a reference, such as std::string& a. But usually these are const. Generally, if you call f(x), if x is an object you should be able to assume that f won't modify x. But if x is a pointer, then you should assume that f might modify the object pointed to by x.)