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Not at all, Rust and C++ are more-or-less equivalent here. In C++, like in Rust, each lambda gets a unique unnamable type, and you can use template parameters t

Not at all, Rust and C++ are more-or-less equivalent here. In C++, like in Rust, each lambda gets a unique unnamable type, and you can use template parameters to express direct calls. > The lambda expression is a prvalue expression of unique unnamed non-union non-aggregate class type, known as closure type, which is declared (for the purposes of ADL) in the smallest block scope, class scope, or namespace scope that contains the lambda expression. https://en.cppreference.com/w/cpp/language/lambda https://en.cppreference.com/w/cpp/language/lambda

tialaramex 3y ago
Of course, but the result is that you need to write a lambda, even when it seems redundant semantically, because otherwise the optimiser may not see what's going on, whereas in Rust the natural expression of what you meant does what you expect. One of the compatibility snags in Rust's standard library is that although you can write "Some words".contains(char::is_whitespace) and that does what you expect, the same won't work for char::is_ascii_whitespace for historic reasons and so you need to write the lambda in that case, which is uglier. Now, in that case if you write the natural thing you get a compiler error at least, but having the unnecessary lambda be needed to get performance is worse. I really thought this was a dig at C++. What were you thinking of?