Question

考虑以下代码：

typedef enum Type1
{
    val11,
    val12
} Type1;

typedef enum Type2
{
    val21,
    val22
} Type2;

Type1 type1 = val11;
if ( type1 == val22 )
    std::cout << "foo";

Visual Studio 2015不会发出任何警告（即使使用/ Wall）。但是type1和val22的类型不同。这是正常的还是Visual Studio错误？

Answer 1

据我所知，编译器没有义务在比较不同类型的枚举时发出警告。我在标准中找不到它。对于经典枚举类型，存在对int的隐式类型转换，因此生成的代码是完全合法的。从语义上来说，比较不同类型的枚举通常是不正确的，因此从C ++开始，我们有一个范围的枚举结构，它不允许隐式转换。（见下面的代码）。

#include <iostream>
using namespace std;

enum UE1 // This is an unscoped enumeration (since C)
{
    val11,
    val12
};

enum UE2 // This is an unscoped enumeration too
{
    val21, // have to use different names for enumeration constants
    val22
};

enum class SE1 // This is an scoped enumeration (since C++11)
{
    val1,
    val2
};

enum class SE2
{
    val1, // can use the same names for the constants
    val2  // because they are in the different scope
};

int main(int, char**)
{
    if (val11 == val22) // implicit conversion from an enum to int is available
        cout << "UE::val11 is equal to UE::val22" << endl;

    if (static_cast<int>(SE1::val1) == static_cast<int>(SE2::val1)) // have to apply explicit conversion
        cout << "SE1::val1 is equal to SE2::val1" << endl;

    if (SE1::val1 == SE2::val1) // error!!! Cannot make implicit conversions from a scoped enumeration.
        cout << "SE1::val1 is equal to SE2::val1" << endl;

    return 0;
}

为什么我没有得到枚举比较不匹配的警告？

1 个答案: