在 C++ 中编译时的 if/else ?

2021-12-12 00:00:00 compilation if-statement templates c++

考虑以下代码:

#include <iostream>
#include <type_traits>

template<typename T> class MyClass
{
    public:
        MyClass() : myVar{0} {;}
        void testIf() {
            if (isconst) {
                myVar;
            } else {
                myVar = 3;
            }
        }
        void testTernary() {
            (isconst) ? (myVar) : (myVar = 3);
        }

    protected:
        static const bool isconst = std::is_const<T>::value;
        T myVar;
};

int main()
{
    MyClass<double> x;
    MyClass<const double> y;
    x.testIf();
    x.testTernary();
    y.testIf(); // <- ERROR
    y.testTernary(); // <- ERROR
    return 0;
}

对于 x(非常量)没有问题.但是 y(const 数据类型)会导致错误，即使 if/else 中的条件在编译时已知.

For x (non-const) there is no problem. But y (const data type) cause an error even if the condition in if/else is known at compile-time.

是否有可能在编译时不编译错误条件?

Is there any possibility to not compile the false condition at compile-time ?

推荐答案

最简单的解决方法是部分模板特化:

The simplest fix is partial template specialization:

template<typename T> class MyClassBase
{
    public:
        MyClassBase() : myVar{0} {;}

    protected:
        T myVar;
};

template<typename T> class MyClass: MyClassBase<T>
{
    public:
        void testIf() { myVar = 3; }
};

template<typename T> class MyClass<const T>: MyClassBase<const T>
{
    public:
        void testIf() { myVar; }
};

另一种选择是委托:

template<typename T> class MyClass
{
    public:
        MyClass() : myVar{0} {;}
        void testIf() { testIf_impl(std::integral_constant<bool, isconst>()); }

    protected:
        static const bool isconst = std::is_const<T>::value;
        T myVar;

    private:
        void testIf_impl(std::true_type) { myvar; }
        void testIf_impl(std::false_type) { myVar = 3; }
};

SFINAE 是另一种选择，但通常不适合这种情况:

SFINAE is another option, but is generally not preferred for this case:

template<typename T> class MyClass
{
    public:
        MyClass() : myVar{0} {;}
        template
        <typename U = void>
        typename std::enable_if<std::is_const<T>::value, U>::type testIf() { myvar; }
        template
        <typename U = void>
        typename std::enable_if<!std::is_const<T>::value, U>::type testIf() { myvar = 3; }

    protected:
        static const bool isconst = std::is_const<T>::value;
        T myVar;
};

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