When can I use explicit operator bool without a cast?

Question

My class has an explicit conversion to bool:

struct T {
    explicit operator bool() const { return true; }
};

and I have an instance of it:

T t;

To assign it to a variable of type bool, I need to write a cast:

bool b = static_cast<bool>(t);
bool b = bool(t);
bool b(t);  // converting initialiser
bool b{static_cast<bool>(t)};

I know that I can use my type directly in a conditional without a cast, despite the explicit qualifier:

if (t)
    /* statement */;

Where else can I use t as a bool without a cast?

Answer 1

The standard mentions places where a value may be "contextually converted to bool". They fall into four main groups:

Statements

•    if (t) /* statement */;
  

•    for (;t;) /* statement */;
  

•    while (t) /* statement */;
  

•    do { /* block */ } while (t);
  

Expressions

•    !t
  

•    t && t2
  

•    t || t2
  

•    t ? "true" : "false"
  

Compile-time tests

•    static_assert(t);
  

•    noexcept(t)
  

•    explicit(t)
  

•    if constexpr (t)
  

The conversion operator needs to be constexpr for these.

Algorithms and concepts

•    NullablePointer T
  

  Anywhere the Standard requires a type satisfying this concept (e.g. the pointer type of a std::unique_ptr), it may be contextually converted. Also, the return value of a NullablePointer's equality and inequality operators must be contextually convertible to bool.

•    std::remove_if(first, last, [&](auto){ return t; });
  

  In any algorithm with a template parameter called Predicate or BinaryPredicate, the predicate argument can return a T.

•    std::sort(first, last, [&](auto){ return t; });
  

  In any algorithm with a template parameter called Compare, the comparator argument can return a T.

(source1, source2)

---

Do be aware that a mix of const and non-const conversion operators can cause confusion:

• Why doesn't explicit bool() conversion happen in contextual conversion?
• Why does the explicit operator bool not in effect as expected?