if statement
Conditionally executes another statement.
Used where code needs to be executed based on a run-time or compile-time condition.
Contents |
[edit] Syntax
attr(optional) if ( condition ) statement-true | (until C++17) | ||||||||
attr(optional) if ( condition ) statement-true else statement-false | (until C++17) | ||||||||
attr(optional) if constexpr(optional) ( init-statement(optional) condition ) statement-true | (since C++17) | ||||||||
attr(optional) if constexpr(optional) ( init-statement(optional) condition ) statement-true else statement-false | (since C++17) | ||||||||
| attr(C++11) | - | any number of attributes |
| condition | - | one of
|
| init-statement(C++17) | - | either
|
| statement-true | - | any statement (often a compound statement), which is executed if condition evaluates to true |
| statement-false | - | any statement (often a compound statement), which is executed if condition evaluates to false |
[edit] Explanation
If the condition yields true after conversion to bool, statement-true is executed.
If the else part of the if statement is present and condition yields false after conversion to bool, statement-false is executed.
In the second form of if statement (the one including else), if statement-true is also an if statement then that inner if statement must contain an else part as well (in other words, in nested if-statements, the else is associated with the closest if that doesn't have an else)
#include <iostream> int main() { // simple if-statement with an else clause int i = 2; if (i > 2) { std::cout << i << " is greater than 2\n"; } else { std::cout << i << " is not greater than 2\n"; } // nested if-statement int j = 1; if (i > 1) if (j > 2) std::cout << i << " > 1 and " << j << " > 2\n"; else // this else is part of if (j > 2), not of if (i > 1) std::cout << i << " > 1 and " << j << " <= 2\n"; // declarations can be used as conditions with dynamic_cast struct Base { virtual ~Base() {} }; struct Derived : Base { void df() { std::cout << "df()\n"; } }; Base* bp1 = new Base; Base* bp2 = new Derived; if (Derived* p = dynamic_cast<Derived*>(bp1)) // cast fails, returns nullptr p->df(); // not executed if (auto p = dynamic_cast<Derived*>(bp2)) // cast succeeds p->df(); // executed}
Output:
2 is not greater than 22 > 1 and 1 <= 2df()
If Statements with InitializerIf 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 statements. std::map<int, std::string> m;std::mutex mx;extern bool shared_flag; // guarded by mxint 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 (auto keywords = {"if", "for", "while"}; std::any_of(keywords.begin(), keywords.end(), [&s](const char* kw) { return s == kw; })) { std::cerr << "Token must not be a keyword\n"; }} | (since C++17) | ||||||||||||||||||||||||||||||||||||||||||||||
Constexpr IfThe statement that begins with In a constexpr if statement, the value of condition must be a contextually converted constant expression of type 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 requiredint f() {if constexpr (true) return 0;else if (x) return x;else return -x;} 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.} 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 }}
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> } };} Note: the discarded statement can't be ill-formed for every possible specialization: template <typename T>void f() { if constexpr (std::is_arithmetic_v<T>) // ... else static_assert(false, "Must be arithmetic"); // ill-formed: invalid for every T} The common workaround for such a catch-all statement is a type-dependent expression that is always false: template<class T> struct dependent_false : std::false_type {};template <typename T>void f() { if constexpr (std::is_arithmetic_v<T>) // ... else static_assert(dependent_false<T>::value, "Must be arithmetic"); // ok} Labels (goto targets, | (since C++17) |
[edit] Notes
If statement_true or statement_false is not a compound statement, it is treated as if it were:
if (x) int i;// i is no longer in scope
is the same as
if (x) { int i;} // i is no longer in scope
The scope of the name introduced by condition, if it is a declaration, is the combined scope of both statements' bodies:
if (int x = f()) { int x; // error: redeclaration of x} else { int x; // error: redeclaration of x}
If statement-true is entered by goto or longjmp, statement_false is not executed. | (since C++14) |
Switch and goto are not allowed to jump into a branch of constexpr if statement. | (since C++17) |
[edit] Keywords
[edit] See Also
| C documentation for if statement |
