explicit specifier doesn't seem to work when converting an object to bool

I am learning C++ recently and I noticed an example on cppreference, part of which goes like this:

struct B

{

    explicit B(int) { }

    explicit operator bool() const { return true; }

};



int main()

{

    B b2(2);       // OK: direct-initialization selects B::B(int)

    if (b2) ;      // OK: B::operator bool()

}

The introduction to implicit conversions tells me "when the expression is used in an if statement or a loop" the result of this expression( b2 ) will be converted into bool type implicitly.

Also, the introduction to explicit specifier tells me if "a conversion function is explicit, it cannot be used for implicit conversions".

Since b2 will be converted implicitly in if(b2), and the conversion function is explicit, how comes if(b2) is ok?

edited Jan 29 at 22:11

Tas

5,51332643

asked Jan 29 at 15:31

Jinlong Chen

664

1

Maybe reated: stackoverflow.com/questions/39995573/…

– Sergey
Jan 29 at 15:35

add a comment |

I am learning C++ recently and I noticed an example on cppreference, part of which goes like this:

struct B

{

    explicit B(int) { }

    explicit operator bool() const { return true; }

};



int main()

{

    B b2(2);       // OK: direct-initialization selects B::B(int)

    if (b2) ;      // OK: B::operator bool()

}

The introduction to implicit conversions tells me "when the expression is used in an if statement or a loop" the result of this expression( b2 ) will be converted into bool type implicitly.

Also, the introduction to explicit specifier tells me if "a conversion function is explicit, it cannot be used for implicit conversions".

Since b2 will be converted implicitly in if(b2), and the conversion function is explicit, how comes if(b2) is ok?

edited Jan 29 at 22:11

Tas

5,51332643

asked Jan 29 at 15:31

Jinlong Chen

664

1

Maybe reated: stackoverflow.com/questions/39995573/…

– Sergey
Jan 29 at 15:35

add a comment |

I am learning C++ recently and I noticed an example on cppreference, part of which goes like this:

struct B

{

    explicit B(int) { }

    explicit operator bool() const { return true; }

};



int main()

{

    B b2(2);       // OK: direct-initialization selects B::B(int)

    if (b2) ;      // OK: B::operator bool()

}

The introduction to implicit conversions tells me "when the expression is used in an if statement or a loop" the result of this expression( b2 ) will be converted into bool type implicitly.

Also, the introduction to explicit specifier tells me if "a conversion function is explicit, it cannot be used for implicit conversions".

Since b2 will be converted implicitly in if(b2), and the conversion function is explicit, how comes if(b2) is ok?

edited Jan 29 at 22:11

Tas

5,51332643

asked Jan 29 at 15:31

Jinlong Chen

664

I am learning C++ recently and I noticed an example on cppreference, part of which goes like this:

struct B

{

    explicit B(int) { }

    explicit operator bool() const { return true; }

};



int main()

{

    B b2(2);       // OK: direct-initialization selects B::B(int)

    if (b2) ;      // OK: B::operator bool()

}

The introduction to implicit conversions tells me "when the expression is used in an if statement or a loop" the result of this expression( b2 ) will be converted into bool type implicitly.

Also, the introduction to explicit specifier tells me if "a conversion function is explicit, it cannot be used for implicit conversions".

Since b2 will be converted implicitly in if(b2), and the conversion function is explicit, how comes if(b2) is ok?

c++ type-conversion explicit

edited Jan 29 at 22:11

Tas

5,51332643

asked Jan 29 at 15:31

Jinlong Chen

664

edited Jan 29 at 22:11

Tas

5,51332643

asked Jan 29 at 15:31

Jinlong Chen

664

edited Jan 29 at 22:11

Tas

5,51332643

edited Jan 29 at 22:11

Tas

5,51332643

edited Jan 29 at 22:11

Tas

5,51332643

asked Jan 29 at 15:31

Jinlong Chen

664

asked Jan 29 at 15:31

Jinlong Chen

664

asked Jan 29 at 15:31

Jinlong Chen

664

1

Maybe reated: stackoverflow.com/questions/39995573/…

– Sergey
Jan 29 at 15:35

add a comment |

1

Maybe reated: stackoverflow.com/questions/39995573/…

– Sergey
Jan 29 at 15:35

Maybe reated: stackoverflow.com/questions/39995573/…

– Sergey
Jan 29 at 15:35

add a comment |

2 Answers
2

active

oldest

votes

Contextual conversion is special; since C++11, explicit conversion functions will be considered in contextual conversions.

(emphasis mine)

(since C++11)

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier; (since C++20)

the predicate of a contract attribute. (since C++20)

That means for if (b2), b2 will be converted to bool implicitly by B::operator bool() even it's declared as explicit.

edited Jan 29 at 16:00

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

add a comment |

Read further in your own link. Contextual conversions occur implicitly even for explicit conversions:

Contextual conversions

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier;

the predicate of a contract attribute.

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Contextual conversion is special; since C++11, explicit conversion functions will be considered in contextual conversions.

(emphasis mine)

(since C++11)

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier; (since C++20)

the predicate of a contract attribute. (since C++20)

That means for if (b2), b2 will be converted to bool implicitly by B::operator bool() even it's declared as explicit.

edited Jan 29 at 16:00

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

add a comment |

Contextual conversion is special; since C++11, explicit conversion functions will be considered in contextual conversions.

(emphasis mine)

(since C++11)

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier; (since C++20)

the predicate of a contract attribute. (since C++20)

That means for if (b2), b2 will be converted to bool implicitly by B::operator bool() even it's declared as explicit.

edited Jan 29 at 16:00

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

add a comment |

Contextual conversion is special; since C++11, explicit conversion functions will be considered in contextual conversions.

(emphasis mine)

(since C++11)

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier; (since C++20)

the predicate of a contract attribute. (since C++20)

That means for if (b2), b2 will be converted to bool implicitly by B::operator bool() even it's declared as explicit.

edited Jan 29 at 16:00

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

Contextual conversion is special; since C++11, explicit conversion functions will be considered in contextual conversions.

(emphasis mine)

(since C++11)

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier; (since C++20)

the predicate of a contract attribute. (since C++20)

That means for if (b2), b2 will be converted to bool implicitly by B::operator bool() even it's declared as explicit.

edited Jan 29 at 16:00

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

edited Jan 29 at 16:00

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

answered Jan 29 at 15:35

songyuanyao

93.9k11182249

add a comment |

Read further in your own link. Contextual conversions occur implicitly even for explicit conversions:

Contextual conversions

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier;

the predicate of a contract attribute.

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

add a comment |

Read further in your own link. Contextual conversions occur implicitly even for explicit conversions:

Contextual conversions

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier;

the predicate of a contract attribute.

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

add a comment |

Read further in your own link. Contextual conversions occur implicitly even for explicit conversions:

Contextual conversions

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier;

the predicate of a contract attribute.

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

Read further in your own link. Contextual conversions occur implicitly even for explicit conversions:

Contextual conversions

In the following contexts, the type bool is expected and the implicit conversion is performed if the declaration bool t(e); is well-formed (that is, an explicit conversion function such as explicit T::operator bool() const; is considered). Such expression e is said to be contextually converted to bool.

the controlling expression of if, while, for;

the operands of the built-in logical operators !, && and ||;

the first operand of the conditional operator ?:;

the predicate in a static_assert declaration;

the expression in a noexcept specifier;

the expression in an explicit specifier;

the predicate of a contract attribute.

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

answered Jan 29 at 15:35

ShadowRanger

63.5k660100

add a comment |

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