Last update August 5, 2009

rvalue

Difference (last change) (Author, normal page display)

Changed: 3c3

As suggested by the definition, mutating operations such as ++, --, +=, *=, =, member function calls, and the like may not be applied to rvalues.

As suggested by the definition, modifying operations such as ++, --, +=, *=, =, member function calls, and the like may not be applied to rvalues.

An expression that cannot have a value assigned to it. The result of calling a function that does not return a reference. [...] Contrast with lvalue.

source

As suggested by the definition, modifying operations such as ++, --, +=, *=, =, member function calls, and the like may not be applied to rvalues.

The dereference operator *expr turns rvalues into lvalues. The reference or address-of operator &expr turns lvalues into rvalues.

Examples:

// D2 code.
class C
{
    int a;
    this() {}
}

struct S
{
    int a;
}

int foo;
int bar() { return foo; }
int* baz() { return &foo; }
ref int qux() { return foo; }
C getC() { return new C(); }
S s;
S getS() { return s; }

void main()
{
    int* ptr = new int;
    
    foo++; // foo is an lvalue because it is a variable.
    bar()++; // Error: bar() is not an lvalue. (It is an rvalue.)
    bar() = 42; // Error: bar() is not an lvalue.  bar() is an rvalue and must appear on the right-hand-side of the assignment.
    baz() = ptr; // Error: baz() is not an lvalue.  A pointer is returned, but not dereferenced.
    *baz() = 42; // Fine.  The dereference operator turns baz() into an lvalue.
    qux() = 42; // Fine.  References are lvalues.

    getC().a = 42; // Fine, getC() is a class instance, and therefore also an lvalue.

    // Compiles, but does nothing.  getS() is not an lvalue, but an rvalue.
    // getS() returns a struct whose a member is set to 42, but the struct is a copy that gets discarded.
    getS().a = 42;

    s.a = 42; // This really will set a to 42.  s is an lvalue because it is a variable.
}