rvalue

If a function returns an int, can it be assigned by an int value? I don't see it makes too much sense to assign a value to a function. int f() {} f() = 1; I noticed that, if the function returns a reference to an int, it is ok. Is it restricted only to int? how about other types? or any other rules? int& f() {} f() = 1; Alexander Gessler The first function returns an integer by-value, which is an r-value . You can't assign to an r-value in general. The second f() returns a reference to an integer, which is a l-value - so you can assign to it. int a = 4, b = 5; int& f() {return a;} ... f() = 6;

Is there a way to write a function in C++ that accepts both lvalue and rvalue arguments, without making it a template? For example, suppose I write a function print_stream that reads from an istream and prints the data that was read to the screen, or something. I think it's reasonable to call print_stream like this: fstream file{"filename"}; print_stream(file); as well as like this: print_stream(fstream{"filename"}); But how do I declare print_stream so that both uses work? If I declare it as void print_stream(istream& is); then the second use won't compile because an rvalue will not bind to a

In C++03, Boost's Foreach, using this interesting technique , can detect at run-time whether an expression is an lvalue or an rvalue. (I found that via this StackOverflow question: Rvalues in C++03 ) Here's a demo of this working at run-time (This is a more basic question that arose while I was thinking about this other recent question of mine . An answer to this might help us answer that other question.) Now that I've spelled out the question, testing rvalue-ness in C++03 at compile-time, I'll talk a little about the things I've been trying so far. I want to be able to do this check at

I am trying to understand why someone would write a function that takes a const rvalue reference . In the code example below what purpose is the const rvalue reference function (returning "3"). And why does overload resolution preference the const Rvalue above the const LValue reference function (returning "2"). #include <string> #include <vector> #include <iostream> std::vector<std::string> createVector() { return std::vector<std::string>(); } //takes movable rvalue void func(std::vector<std::string> &&p) { std::cout << "1"; } //takes const lvalue void func(const std::vector<std::string> &p)

#include <stdio.h> int main() { int i = 10; printf("%d\n", ++(-i)); // <-- Error Here } What is wrong with ++(-i) ? Please clarify. -i generates a temporary and you can't apply ++ on a temporary(generated as a result of an rvalue expression). Pre increment ++ requires its operand to be an lvalue, -i isn't an lvalue so you get the error. The ++ operator increments a variable. (Or, to be more precise, an lvalue —something that can appear on the l eft side of an assignment expression) (-i) isn't a variable, so it doesn't make sense to increment it. Lee Louviere You can't increment a temporary