WebOct 1, 2008 · std::cout << typeid( A ).name() << std::endl ;} {class A {} ; std::cout << typeid( A ).name() << std::endl ;}} void f() {g() ;} (Note that the definitions of class A in g() are at exactly the same line in both files---VC++ encodes the line number of a local class in its name, so they only have the same name if they are defined on the same line.)-- WebCreated attachment 33561 sample code which reproduces the bug a c++11 lambda expression `[]()->void{}` defined in the `main` function gets `Z4mainEUlvE_` as typeid.name which can be demangled to `main::{lambda()#1}` using `abi::__cxa_demangle`. However a globally defined lambda gets reported as just `UlvE_` and the demangle function fails …
typeid operator - cppreference.com
Weba) If expression is an lvalue (until C++11) a glvalue (since C++11) expression that identifies an object of a polymorphic type (that is, a class that declares or inherits at least one … WebAs others have said, the result here is implementation-defined, meaning that the implementation (i.e., the compiler toolchain) is free to define it how it wants, so long as it … recycling and refuse
C++报错:expected type-specifier before ‘QSrialPort‘ - CSDN博客
WebYou can retrieve the implementation defined name of a type in runtime by using the .name () member function of the std::type_info object returned by typeid. #include #include int main () { int speed = 110; std::cout << typeid (speed).name () << '\n'; } Output (implementation-defined): int. WebSep 17, 2024 · 今天在看代码中,看到了一个很不错的工具函数typeid().name(),可以用来返回变量的数据类型,很实用。下面来具体学习一下该函数。 首先来看typeid操作符,其 … WebC++ 模板化函数不接受基元类型?,c++,templates,C++,Templates,这段代码没有编译,但我不知道为什么,而且typeid函数可以将int作为输入参数,所以问题一定与模板机制有关,但我不知道失败的原因 #include #include template void func(T) { std::cout << typeid(T).name() << std::endl; } int main() { func(int); recycling and reusing