< cpp‎ | utility
Utilities library
Language support
Type support (basic types, RTTI)
Library feature-test macros (C++20)
Dynamic memory management
Program utilities
Coroutine support (C++20)
Variadic functions
Debugging support
Three-way comparison
General utilities
Date and time
Function objects
Formatting library (C++20)
Relational operators (deprecated in C++20)
Integer comparison functions
Swap and type operations
Common vocabulary types
Elementary string conversions

Dynamic memory management
Uninitialized memory algorithms
Constrained uninitialized memory algorithms
Garbage collection support
(C++11)(until C++23)
(C++11)(until C++23)
(C++11)(until C++23)
(C++11)(until C++23)
(C++11)(until C++23)
(C++11)(until C++23)

Defined in header <new>
template< class T >
constexpr T* launder( T* p ) noexcept;
(since C++17)

Provenance fence with respect to p. Returns a pointer to the same memory that p points to, but where the referent object is assumed to have a distinct lifetime and dynamic type.

Formally, given

  • the pointer p represents the address A of a byte in memory
  • an object x is located at the address A
  • x is within its lifetime
  • the type of x is the same as T, ignoring cv-qualifiers at every level
  • every byte that would be reachable through the result is reachable through p (bytes are reachable through a pointer that points to an object y if those bytes are within the storage of an object z that is pointer-interconvertible with y, or within the immediately enclosing array of which z is an element).

Then std::launder(p) returns a value of type T* that points to the object x. Otherwise, the behavior is undefined.

The program is ill-formed if T is a function type or (possibly cv-qualified) void.

std::launder may be used in a core constant expression if and only if the (converted) value of its argument may be used in place of the function invocation. In other words, std::launder does not relax restrictions in constant evaluation.

[edit] Notes

std::launder has no effect on its argument. Its return value must be used to access the object. Thus, it's always an error to discard the return value.

Typical uses of std::launder include:

  • Obtaining a pointer to an object created in the storage of an existing object of the same type, where pointers to the old object cannot be reused (for instance, because either object is a base class subobject);
  • Obtaining a pointer to an object created by placement new from a pointer to an object providing storage for that object.

The reachability restriction ensures that std::launder cannot be used to access bytes not accessible through the original pointer, thereby interfering with the compiler's escape analysis.

int x[10];
auto p = std::launder(reinterpret_cast<int(*)[10]>(&x[0])); // OK
int x2[2][10];
auto p2 = std::launder(reinterpret_cast<int(*)[10]>(&x2[0][0]));
// Undefined behavior: x2[1] would be reachable through the resulting pointer to x2[0]
// but is not reachable from the source
struct X { int a[10]; } x3, x4[2]; // standard layout; assume no padding
auto p3 = std::launder(reinterpret_cast<int(*)[10]>(&x3.a[0])); // OK
auto p4 = std::launder(reinterpret_cast<int(*)[10]>(&x4[0].a[0]));
// Undefined behavior: x4[1] would be reachable through the resulting pointer to x4[0].a
// (which is pointer-interconvertible with x4[0]) but is not reachable from the source
struct Y { int a[10]; double y; } x5;
auto p5 = std::launder(reinterpret_cast<int(*)[10]>(&x5.a[0]));
// Undefined behavior: x5.y would be reachable through the resulting pointer to x5.a
// but is not reachable from the source

[edit] Example

#include <cassert>
#include <cstddef>
#include <new>
struct Base
    virtual int transmogrify();
struct Derived : Base
    int transmogrify() override
        new(this) Base;
        return 2;
int Base::transmogrify()
    new(this) Derived;
    return 1;
static_assert(sizeof(Derived) == sizeof(Base));
int main()
    // Case 1: the new object failed to be transparently replaceable because
    // it is a base subobject but the old object is a complete object.
    Base base;
    int n = base.transmogrify();
    // int m = base.transmogrify(); // undefined behavior
    int m = std::launder(&base)->transmogrify(); // OK
    assert(m + n == 3);
    // Case 2: access to a new object whose storage is provided
    // by a byte array through a pointer to the array.
    struct Y { int z; };
    alignas(Y) std::byte s[sizeof(Y)];
    Y* q = new(&s) Y{2};
    const int f = reinterpret_cast<Y*>(&s)->z; // Class member access is undefined
                                               // behavior: reinterpret_cast<Y*>(&s)
                                               // has value "pointer to s" and does
                                               // not point to a Y object
    const int g = q->z; // OK
    const int h = std::launder(reinterpret_cast<Y*>(&s))->z; // OK
    [](...){}(f, g, h); // evokes [[maybe_unused]] effect

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

DR Applied to Behavior as published Correct behavior
LWG 2859 C++17 definition of reachable did not consider pointer
arithmetic from pointer-interconvertible object
LWG 3495 C++17 std::launder might make pointer to an inactive
member dereferenceable in constant expression