Last update August 5, 2009

LValue

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

Deleted: 1,50d0

An expression that represents an object or a function and that can be both examined and altered. Contrast with rvalue.

source

Typically an lvalue is a variable, returned reference, pointer being dereferenced, indexed array element, or class instance (Object).

Mutating operations like ++, --, +=, *=, =, member function calls, and such may be applied to lvalues.

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

Examples:
[[code]
// 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.
}
]