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  2. Compatibility of C and C++ - Wikipedia

    en.wikipedia.org/wiki/Compatibility_of_C_and_C++

    C++ is also more strict in conversions to enums: ints cannot be implicitly converted to enums as in C. Also, enumeration constants (enum enumerators) are always of type int in C, whereas they are distinct types in C++ and may have a size different from that of int. [needs update] In C++ a const variable must be initialized; in C this is not ...

  3. Enumerated type - Wikipedia

    en.wikipedia.org/wiki/Enumerated_type

    The Java standard library provides utility classes to use with enumerations. The EnumSet class implements a Set of enum values; it is implemented as a bit array, which makes it very compact and as efficient as explicit bit manipulation, but safer. The EnumMap class implements a Map of enum values to object. It is implemented as an array, with ...

  4. C++11 - Wikipedia

    en.wikipedia.org/wiki/C++11

    enum Enum1; // Invalid in C++03 and C++11; the underlying type cannot be determined. enum Enum2: unsigned int; // Valid in C++11, the underlying type is specified explicitly. enum class Enum3; // Valid in C++11, the underlying type is int. enum class Enum4: unsigned int; // Valid in C++11. enum Enum2: unsigned short; // Invalid in C++11 ...

  5. C++ - Wikipedia

    en.wikipedia.org/wiki/C++

    C++ has enumeration types that are directly inherited from C's and work mostly like these, except that an enumeration is a real type in C++, giving added compile-time checking. Also (as with structs), the C++ enum keyword is combined with a typedef, so that instead of naming the type enum name, simply name it name.

  6. Tagged union - Wikipedia

    en.wikipedia.org/wiki/Tagged_union

    The enum types in the Rust, Haxe, and Swift languages also work as tagged unions. The variant library from the Boost C++ Libraries demonstrated it was possible to implement a safe tagged union as a library in C++, visitable using function objects.

  7. typename - Wikipedia

    en.wikipedia.org/wiki/Typename

    This code looks like it should compile, but it is incorrect because the compiler does not know if T::bar is a type or a value. The reason it doesn't know is that T::bar is a "template-parameter dependent name", or "dependent name" for short, which then could represent anything named "bar" inside a type passed to foo(), which could include typedefs, enums, variables, etc.

  8. static_cast - Wikipedia

    en.wikipedia.org/wiki/Static_cast

    converting a pointer of a base class to a pointer of a non-virtual derived class (downcasting); converting numeric data types such as enums to ints or floats . Although static_cast conversions are checked at compile time to prevent obvious incompatibilities, no run-time type checking is performed that would prevent a cast between incompatible ...

  9. Comparison of Java and C++ - Wikipedia

    en.wikipedia.org/wiki/Comparison_of_Java_and_C++

    Another way is to make another class that extends java.lang.Enum<E>) and may therefore define constructors, fields, and methods as any other class. As of C++11, C++ supports strongly-typed enumerations which provide more type-safety and explicit specification of the storage type.