Report a bug
If you spot a problem with this page, click here to create a Bugzilla issue.
Improve this page
Quickly fork, edit online, and submit a pull request for this page. Requires a signed-in GitHub account. This works well for small changes. If you'd like to make larger changes you may want to consider using a local clone.

Coralling Wild Pointers With ref return scope

A wild pointer is a pointer that escapes its corral, or in other words, escapes its scope. A pointer within its scope is valid, and outside its scope is not valid. Attempting to read or write using an out-of-scope pointer will produce undefined behavior. Undefined behavior can lead to crashes, corruption, malware, and other costly problems. D uses ref, return and scope keywords to prevent pointer escapes.

What is a Scope?

The scope of a declaration is closely related to its lifetime. Thread local variables have a lifetime that's the life of the thread they are in. Global variables have a lifetime from the program start to its finish. A local variable has a lifetime from its initialization to its closing curly brace.

Note on examples: @safe annotation is assumed

int x; // thread local lifetime
__gshared int y; // global lifetime

void mars(int i /* lifetime of i is from function call to function return */)
    { // open new scope
      int* q = &i;  // lifetime starts after q is set to the address of i
      *q = 3;       // sets i to 3
    }               // q's lifetime ends with the end of the scope
    *q = 4;         // oops, can't use q here
    int* p = &i;    // lifetime starts after p is set to the address of i
    *p = 5;         // sets i to 5
} // lifetimes of i and p end when function returns

What is an Escaping Pointer?

A pointer escapes when its value becomes available outside the scope of the pointer. An example of an escaping pointer:

int* escape()
    int i;
    int* p = &i; // create pointer to local variable
    return p;    // return pointer to a local variable that is no longer live

void crash()
    int* q = escape();
    *q = 5;  // unleash the Hounds of Hell

Pointer escapes can occur in many, often not-so-obvious, cases. The compiler is the perfect tool to detect all the cases and report them as errors. Even if a particular pointer escape is benign, if a function interface makes clear that a function arguments cannot escape, it improves user understanding of the function.

The Role of scope

scope is a storage class. When it is applied to a pointer variable, then the pointer's value is mechanically (i.e. enforced by the compiler) prevented from outlasting the scope of the pointer variable. The previous example is modified with the addition of scope:

int* escape()
    int i;
    scope int* p = &i; // p is a scoped pointer
    return p;    // error: scoped pointer p is escaping
The compiler, for this case, helpfully goes one better:
int* escape()
    int i;
    int* p = &i; // p is inferred to be a scoped pointer
    return p;    // error: scoped pointer p is escaping

I.e. if a pointer variable is set to be the address of a local variable, or to the contents of scope pointer, then that pointer variable is automatically set to be a scope pointer. The compiler is pretty good at inferring scope, thus relieving the programmer of adding a lot of annotations.

Note that as scope is a storage class, not a type constructor, it is not possible to specify a scope pointer to a scope pointer. It surprisingly turns out to not be necessary to support that.

The Role of return scope

Consider the following:

void f()
    int i;
    *process(&i) = 4;

int* process(scope int* p) { return p; }

This is perfectly legitimate code, there is no pointer escaping bug. But it won't compile. Without scope on the parameter p, the call to process(&i) would be disallowed. But with scope on p, the return p; is disallowed.

The solution is adding a return annotation:

int* process(return scope int* p) { return p; }

which allows the scope pointer value to be returned by the function.

If a function returns void, return scope also allows returning the scope value through the first parameter:

void mun(ref int* v, return scope int* p)
    v = p; // ok

And that's it for pointers. Remember that pointer scoping is concerned with the value of the pointer variable.

The Role of ref

A ref is a reference to a value, a fancy way of representing a pointer to a value. It is distinguished from a pointer:

  1. by not allowing arithmetic on the address
  2. whenever the ref variable is used an automatic dereference is performed
  3. a ref cannot escape from a function
ref int fin(ref int i)
    return i; // error, cannot return ref variable i by ref

The role of return ref

But it will be allowed if return is applied:

ref int fin(return ref int i)
    return i; // ok

The role of ref scope

The storage classes ref and scope together are orthogonal, they do not affect each other. ref refers to the address of the variable, scope refers to the contents of the variable.

The role of return ref scope

The return here applies to the ref, not the scope.

Let's try tricking the compiler:

ref int* fin(return ref scope int* p) { return p; } // ok

int* tricky()
    int i;
    int* p = &i; // p is now inferred to be scope
    auto q = fin(p);  // q now contains the address of i, and so scope is also inferred
    return q;  // error: scope variable `q` may not be returned

Curses! Foiled again!

The operational idea here is, while compiling @safe code, it is not be possible to escape a scoped value, no matter how twisty the code is.