如何在Xcode调试器中查看推导出的模板类型？

Question

如何在Xcode调试器中查看推断出的模板类型？

使用Meyers＆＃39;中第1项的例子。 Effective Modern C ++（2015），注意变量的值及其类型在Xcode＆＃34;变量视图中是如何显示的，＆＃34;但是没有办法看到变量T的结果推论。

是否可以在Xcode调试器中看到推导出的T类型？

我的老板给了我typeId，如下所示，但我希望直接在调试器中看到该类型。此外，此命令会在IDE之间产生不一致的结果;例如，Xcode输出＆＃34; i＆＃34;对于int，而Visual Studio输出＆＃34; int。＆＃34;

完整的可执行文件和输出如下所示。 Thx，Keith：^）

Xcode屏幕截图

最小，完整且可验证的示例：

#include <iostream>

template <typename T>
int case1a(T& param) {
    return param;
}
template <typename T>
int case1b(const T& param) {
    std::cout << "Type ID   : " << typeid(T).name() << std::endl;
    std::cout << "Type param: " << typeid(param).name() << std::endl;
    return param;
}
template <typename T>
int case2(T&& param) {
    return param;
}
template <typename T>
int case3(T param) {
    return param;
}
int main() {

    int a(23);
    const int ar(a);
    const int & car = a;

    float b(7.);
    std::cout << "case1a(a): " << case1a(a) << std::endl;
    std::cout << "case1a(ar): " << case1a(ar) << std::endl;
    std::cout << "case1a(car): " << case1a(car) << std::endl;

    std::cout << "case1b(a): " << case1b(a) << std::endl;
    std::cout << "case1b(ar): " << case1b(ar) << std::endl;
    std::cout << "case1b(car): " << case1b(car) << std::endl;
    std::cout << "case1b(b): " << case1b(b) << std::endl;

    std::cout << "case2(a): " << case2(a) << std::endl;
    std::cout << "case2(ar): " << case2(ar) << std::endl;
    std::cout << "case2(car): " << case2(car) << std::endl;

    std::cout << "case3(a): " << case3(a) << std::endl;
    std::cout << "case3(ar): " << case3(ar) << std::endl;
    std::cout << "case3(car): " << case3(car) << std::endl;

    std::cout << std::endl;
    return 0;
}

输出

case1a(a): 23
case1a(ar): 23
case1a(car): 23
case1b(a): Type ID   : i
Type param: i
23
case1b(ar): Type ID   : i
Type param: i
23
case1b(car): Type ID   : i
Type param: i
23
case1b(b): Type ID   : f
Type param: f
7
case2(a): 23
case2(ar): 23
case2(car): 23
case3(a): 23
case3(ar): 23
case3(car): 23

Program ended with exit code: 0