标签: is-same

I was trying to static_assert that some meta transformer algorithm worked, and it incredibly did not compare to same, even though the typeid().name() returned the exact same string.

A repetition of the type expression in the typedef could fix the is_same, but I can't understand how repeating the initializer expression in the typedef changes the type, over taking the decltype of an auto variable initialized with that same expression.

A more concrete explanation of what I was doing: …

在使用 clang 编译器的 C++17 中，无论我这样做，我都会遇到相同的构建错误：

EXPECT_TRUE(std::is_same_v<decltype(var1), decltype(var2)>);

或这个：

EXPECT_TRUE(typename std::is_same_v<decltype(var1), decltype(var2)>);,

或这个：

EXPECT_TRUE(typename std::is_same_v<typename decltype(var1), typename decltype(var2)>);

构建命令：

bazel test //my_target_dir:my_target

构建错误：

error: too many arguments provided to function-like macro invocation
                decltype(var2)>);
                ^
gtest/gtest.h:1980:9: note: macro 'EXPECT_TRUE' defined here
#define EXPECT_TRUE(condition) \
        ^
myfile.cpp:125:5: error: use of undeclared identifier 'EXPECT_TRUE'
    EXPECT_TRUE(std::is_same_v<
    ^

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请注意，Googletest 的定义EXPECT_TRUE()在这里：https://github.com/google/googletest/blob/master/googletest/include/gtest/gtest.h#L1980。

我正在做的事情有什么问题，我怎样才能编译它？

参考：

1. std::is_same<T, U>::value和std::is_same_v<T, U>
2. GoogleTest（gtest）文档

如何检查两个类模板实例化是否属于同一个类模板。这是我的代码

#include <iostream>
#include <type_traits>

template<typename T1, typename T2>
class A {
    float val;
public:

};

int main() {
    A<double, double> a_float_type;
    A<int, int> a_int_type;

    // how to check whether both a_double_type and a_int_type were instantiated from the same template class A<,>
    std::cout << std::is_same<decltype(a_float_type), decltype(a_int_type)>::value << std::endl; // returns false
}

我的编译器只支持C++11

考虑我有结构 RGB 和 ARGB。

template<typename T>
struct RGB {
    T r,g,b;
};

template<typename T>
struct ARGB {
    T a,r,g,b;
}

现在我通过以下来定义它们。

using RGB_888 = RGB<unsigned char>;
using RGB_Float = RGB<float>;
using ARGB_8888 = ARGB<unsigned char>;
using ARGB_Float = ARGB<float>;

在某些时候，我想从一个 rgb 转换为另一个，从 rgb 转换为 argb。所以我做以下事情。

template<typename Source, typename Dest>
void convert(const Source& source, Dest& dest)
{

}

它会像这样工作。

RGB_888 rgb888{123,22,311};
RGB_Float rgbFloat;
convert<RGB_888,RGB_Float>(rgb888,rgbFloat);

RGB_888 rgb888(123,22,132};
ARGB_Float argbFloat;
convert<RGB_888,ARGB_Float>(rgb888,argbFloat);

问题是我无法检测typename Source和typename Dest是否来自相同的颜色模型，我需要正确转换。换句话说，如果一些typename …