Best Python templating library to facilitate code generation [closed]
Best suggestion: try them all. It won’t take long. My favourite: Jinja2 (by a mile) It has decent syntax, can trace errors through it, and is sandboxable.
Best suggestion: try them all. It won’t take long. My favourite: Jinja2 (by a mile) It has decent syntax, can trace errors through it, and is sandboxable.
obj.bar<double>(1,2); // This line is faulty. The template keyword is required here, as obj is an instance of a type Foo<T> which depends on the template parameter T, and so the above should be written as: obj.template bar<double>(1,2); //This line is corrected ๐ Read @Johannes’s answer here for detail explanation: Where and why do I … Read more
A conditional base class may be used: struct BaseWithVariable { int addedVariable; }; struct BaseWithoutVariable { }; template <bool AddMembers> class MyClass : std::conditional<AddMembers, BaseWithVariable, BaseWithoutVariable>::type { // etc. };
If we ask the compiler to tell us anything about a class type T that has not even been declared we are bound to get a compilation error. There is no way around that. Therefore if we want to know whether class T “exists”, where T might not even have been declared yet, we must … Read more
In your first example, static_assert should take a second parameter which would be a string literal, otherwise it’s deemed to fail (edit: dropping the the second parameter is legal since C++17). And this second argument cannot be defaulted. Your second example is incorrect for several reasons: decltype is meant to be used on an expression, … Read more
In a nutshell, auto cannot be used in an effort to omit the actual types of function arguments, so stick with function templates and/or overloads. auto is legally used to automatically deduce the types of variables: auto i=5; Be very careful to understand the difference between the following, however: auto x=… auto &x=… const auto … Read more
Which part of the standard says that this is wrong? That would be ยง14.3.1/2 from the 2003 C++ Standard: A local type, a type with no linkage, an unnamed type or a type compounded from any of these types shall not be used as a template-argument for a template type-parameter. How can make this code … Read more
void foo(int… args) {} No you cannot write that. But you can have the same effect with this approach: template<typename …Ints> void foo(Ints… ints) { int args[] { ints… }; //unpack ints here //use args } With this approach, you can pass all int if you want. If any argument passed to foo is not … Read more