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A constexpr lambda is a lambda expression whose invocation can be evaluated during compile-time constant evaluation. Introduced in C++17, it allows the compiler to declare the generated closure type’s operator() as a constexpr member function, enabling the lambda to participate in constant expressions.

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Syntax

The constexpr specifier is placed after the parameter list and mutable specifier (if any), but before the trailing return type and body. 
auto lambda = [captures](parameters) constexpr -> return_type {
    // body
};

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Implicit vs. Explicit constexpr

In C++17 and later, lambdas are implicitly constexpr. The compiler automatically treats the lambda’s operator() as constexpr if its body satisfies the requirements of a constant expression. Providing the constexpr keyword explicitly declares the operator() as constexpr. Prior to C++23, this mandated that the lambda must be valid in at least one constant expression context. As of C++23 (via proposal P2448R2), this strict diagnostic requirement is removed. The compiler will accept the explicit constexpr definition even if no valid constant expression context exists, though the function still cannot be evaluated at compile-time if the evaluated path executes non-constexpr operations. 
// Implicit constexpr: Valid in constant expressions
auto implicit_add = [](int a, int b) { return a + b; };
static_assert(implicit_add(2, 3) == 5); 

// Explicit constexpr
auto explicit_square = [](int x) constexpr { return x * x; };
constexpr int val = explicit_square(4);

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Technical Requirements

For a lambda’s invocation to be successfully evaluated at compile-time, its generated operator() must adhere to standard constexpr function constraints during evaluation:
  1. Literal Types: All parameters and return types must be literal types.
  2. Valid Operations: The execution path evaluated at compile-time cannot perform operations invalid in a constant expression, such as calling non-constexpr functions or executing inline assembly.
  3. Control Flow and State (C++23 Relaxations): Prior to C++23, goto statements, labels, and definitions of static or thread_local variables were strictly prohibited within the lambda body. As of C++23 (via proposal P2242R3), these constructs are permitted in the definition, provided the specific control flow path evaluated during constant evaluation does not actually execute or initialize them.
  4. Memory Allocation: In C++20 and later, transient constexpr allocations (new/delete) are permitted provided the memory is dynamically allocated and deallocated within the same constant evaluation context.

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Capture Semantics in Constant Expressions

When a lambda captures variables, the resulting closure object contains data members corresponding to those captures. For a stateful lambda to be invoked in a constant expression:
  • The captured variables must be literal types.
  • The closure object variable itself must be declared constexpr so that it is usable in constant expressions. Merely initializing the closure object with a constant expression is insufficient if the closure has state.
  • If capturing by reference, the referenced object must have static storage duration or otherwise be usable in a constant expression.
constexpr int multiplier = 10;

// The closure object itself must be declared constexpr 
// to be usable in the static_assert below.
constexpr auto multiply = [multiplier](int x) constexpr {
    return x * multiplier;
};

static_assert(multiply(5) == 50); // Valid

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C++20 Addition: consteval Lambdas

While constexpr implies the lambda can be evaluated at compile-time, C++20 introduced the consteval specifier for lambdas to create immediate functions. A consteval lambda must be evaluated at compile-time; any attempt to invoke it in a runtime context results in a compilation error. 
auto strict_compile_time = [](int x) consteval { 
    return x * 2; 
};

constexpr int a = strict_compile_time(10); // OK
int b = 5;
// int c = strict_compile_time(b);         // Error: 'b' is not a constant expression
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