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if statement

Conditionally executes another statement.

Used where code needs to be executed based on a run-time or compile-time condition.

Syntax

attr(optional) if ( condition ) statement-true (1)
attr(optional) if ( condition ) statement-true else statement-false (2)
(until C++17)
attr(optional) if constexpr(optional) ( init-statement(optional) condition ) statement-true (1)
attr(optional) if constexpr(optional) ( init-statement(optional) condition ) statement-true else statement-false (2)
(since C++17)
attr(C++11) - any number of attributes
condition - one of
init-statement(C++17) - either
  • an expression statement (which may be a null statement ";")
  • a simple declaration, typically a declaration of a variable with initializer, but it may declare arbitrary many variables or be a decomposition declaration
Note that any init-statement must end with a semicolon ;, which is why it is often described informally as an expression or a declaration followed by a semicolon.
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

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 NULL
       p->df();  // not executed
 
   if (auto p = dynamic_cast<Derived*>(bp2)) // cast succeeds
       p->df();  // executed
}

Output:

2 is not greater than 2
2 > 1 and 1 <= 2
df()

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

{
init_statement
if constexpr(optional) ( condition )
statement-true

}

or.

{
init_statement
if constexpr(optional) ( condition )
statement-true
else statement-false

}

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 mx
int demo() {
   if (auto it = m.find(10); it != m.end()) { return it->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 If

The statement that begins with if constexpr is known as the constexpr if statement.

In a constexpr if statement, the value of condition must be a contextually converted constant expression of type bool. If the value is 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, and 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 (in particular, it can odr-use a variable that is not defined).

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

Note: an example where the condition remains value-dependent after instantiation is a nested template, e.g.

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>
        }
    };
}
 (since C++17)

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)

Keywords

if, else, constexpr.

See Also

C documentation for if statement

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