if statement

if statements with initializer

If init-statement is used, the if statement is equivalent to

or

Except that names declared by the init-statement (if init-statement is a declaration) and names declared by condition (if condition is a declaration) are in the same scope, which is also the scope of both statement s.

std::map<int, std::string> m;
std::mutex mx;
extern bool shared_flag; // guarded by mx
 
int demo()
{
    if (auto it = m.find(10); it != m.end())
        return it->second.size();
 
    if (char buf[10]; std::fgets(buf, 10, stdin))
        m[0] += buf;
 
    if (std::lock_guard lock(mx); shared_flag)
    {
        unsafe_ping();
        shared_flag = false;
    }
 
    if (int s; int count = ReadBytesWithSignal(&s))
    {
        publish(count);
        raise(s);
    }
 
    if (const auto keywords = {"if", "for", "while"};
        std::ranges::any_of(keywords, [&tok](const char* kw) { return tok == kw; }))
    {
        std::cerr << "Token must not be a keyword\n";
    }
}

Constexpr if

The statement that begins with if constexpr is known as the constexpr if statement. All substatements of a constexpr if statement are control-flow-limited statements.

In a constexpr if statement, condition must be a contextually converted constant expression of type bool(until C++23)an expression contextually converted to bool, where the conversion is a constant expression(since C++23).

If condition yields true, then statement-false is discarded (if present), otherwise, statement-true is discarded.

The return statements in a discarded statement do not participate in function return type deduction:

template<typename T>
auto get_value(T t)
{
    if constexpr (std::is_pointer_v<T>)
        return *t; // deduces return type to int for T = int*
    else
        return t;  // deduces return type to int for T = int
}

The discarded statement can ODR-use a variable that is not defined:

extern int x; // no definition of x required
 
int f()
{
    if constexpr (true)
        return 0;
    else if (x)
        return x;
    else
        return -x;
}

Outside a template, a discarded statement is fully checked. if constexpr is not a substitute for the #if preprocessing directive:

void f()
{
    if constexpr(false)
    {
        int i = 0;
        int *p = i; // Error even though in discarded statement
    }
}

If a constexpr if statement appears inside a templated entity, and if condition is not value-dependent after instantiation, the discarded statement is not instantiated when the enclosing template is instantiated.

template<typename T, typename ... Rest>
void g(T&& p, Rest&& ...rs)
{
    // ... handle p
    if constexpr (sizeof...(rs) > 0)
        g(rs...); // never instantiated with an empty argument list
}

The condition remains value-dependent after instantiation is a nested template:

template<class T>
void g()
{
    auto lm = [=](auto p)
    {
        if constexpr (sizeof(T) == 1 && sizeof p == 1)
        {
            // this condition remains value-dependent after instantiation of g<T>,
            // which affects implicit lambda captures
            // this compound statement may be discarded only after
            // instantiation of the lambda body
        }
    };
}

The discarded statement cannot be ill-formed for every possible specialization:

template<typename T>
void f()
{
    if constexpr (std::is_arithmetic_v<T>)
        // ...
    else {
        using invalid_array = int[-1]; // ill-formed: invalid for every T
        static_assert(false, "Must be arithmetic"); // ill-formed before CWG2518
    }
}

The common workaround before the implementation of CWG issue 2518 for such a catch-all statement is a type-dependent expression that is always false:

template<typename>
inline constexpr bool dependent_false_v = false;
 
template<typename T>
void f()
{
    if constexpr (std::is_arithmetic_v<T>)
        // ...
    else {
        // workaround before CWG2518
        static_assert(dependent_false_v<T>, "Must be arithmetic");
    }
}

A typedef declaration or alias declaration(since C++23) can be used as the init-statement of a constexpr if statement to reduce the scope of the type alias.

Consteval if

The statement that begins with if consteval is known as the consteval if statement. All substatements of a consteval if statement are control-flow-limited statements.

statement must be a compound statement, and it will still be treated as a part of the consteval if statement even if it is not a compound statement (and thus results in a compilation error):

constexpr void f(bool b)
{
    if (true)
        if consteval {}
        else ; // error: not a compound-statement
               // else not associated with outer if
}

If a consteval if statement is evaluated in a manifestly constant-evaluated context, compound-statement is executed. Otherwise, statement is executed if it is present.

If the statement begins with if !consteval, the compound-statement and statement (if any) must both be compound statements. Such statements are not considered consteval if statements, but are equivalent to consteval if statements:

• if !consteval {/* stmt */} is equivalent to if consteval {} else {/* stmt */}.
• if !consteval {/* stmt-1 */} else {/* stmt-2 */} is equivalent to if consteval {/* stmt-2 */} else {/* stmt-1 */}.

compound-statement in a consteval if statement (or statement in the negative form) is in an immediate function context, in which a call to an immediate function needs not to be a constant expression.

#include <cmath>
#include <cstdint>
#include <cstring>
#include <iostream>
 
constexpr bool is_constant_evaluated() noexcept
{
    if consteval { return true; } else { return false; }
}
 
constexpr bool is_runtime_evaluated() noexcept
{
    if not consteval { return true; } else { return false; }
}
 
consteval std::uint64_t ipow_ct(std::uint64_t base, std::uint8_t exp)
{
    if (!base) return base;
    std::uint64_t res{1};
    while (exp)
    {
        if (exp & 1) res *= base;
        exp /= 2;
        base *= base;
    }
    return res;
}
 
constexpr std::uint64_t ipow(std::uint64_t base, std::uint8_t exp)
{
    if consteval // use a compile-time friendly algorithm
    {
        return ipow_ct(base, exp);
    }
    else // use runtime evaluation
    {
        return std::pow(base, exp);
    }
}
 
int main(int, const char* argv[])
{
    static_assert(ipow(0, 10) == 0 && ipow(2, 10) == 1024);
    std::cout << ipow(std::strlen(argv[0]), 3) << '\n';
}