Error Invalid Cast Of An Rvalue Expression Of Type
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here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers lvalue and rvalue in c or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question rvalue reference c++ x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; difference between lvalue and rvalue in c it only takes a minute: Sign up Can't understand compiler's behaviour (creating object) up vote 2 down vote favorite I wrote the following code: #include
Lvalue And Rvalue Error In C
std::cout << "copied" << std::endl; } A& get(){ std::cout <<"got" << std::endl; return *this; } ~A(){ std::cout << "destroyed" << std::endl; } }; Now, lines A a = A().get(); and A a; a = A(); compile and work correctly, but A a = A(); claims: no matching function for call to ‘A::A(A)’ note: candidates are: A::A(A&) note: A::A() And to make things explicit, A a = (A&)A(); claims: error: invalid cast of an rvalue expression of type ‘A’ to type lvalue required error in c ‘A&’ I completely don't understand this behaviour. P.S. I know, that if I make const reference in copy c_tor, everything will be OK. c++ share|improve this question edited Oct 22 '11 at 16:38 Raphael R. 8,67611316 asked Oct 22 '11 at 15:01 Lol4t0 10.5k21647 add a comment| 3 Answers 3 active oldest votes up vote 2 down vote accepted A a = A(); This line tries to call the copy-constructor passing a temporary object, and the copy-constructor takes argument by non-const reference. But a temporary object cannot be bound to a non-const reference. That is why you get the compilation error. However, a temporary object can be bound to const reference. So, the solution is to make the parameter const reference as: A(const A& other) { std::cout << "copied" << std::endl; } share|improve this answer answered Oct 22 '11 at 15:06 Nawaz 203k62447655 1 this explanation sounds most logical for me –Lol4t0 Oct 22 '11 at 17:54 add a comment| up vote 3 down vote The copy constructor should get its argument as a const reference (or a simple A value). In your current setup, the program would have to make a mutable reference to a temporary, which is invalid. share|improve this answer answered Oct 22 '11 at 15:03 Fred Foo 228k34429605 You know, g++ does not think so sometimes. For exasmple, private member of type std::auto_ptr
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What Is Lvalue In C
Projects 0 Pulse Graphs New issue fail build on linux #883 Closed sl1pkn07 opened this
Rvalue And Lvalue In Java
Issue Dec 16, 2015 · 19 comments Projects None yet Labels None yet Milestone No milestone Assignees No one assigned 11 http://stackoverflow.com/questions/7860432/cant-understand-compilers-behaviour-creating-object participants sl1pkn07 commented Dec 16, 2015 with GCC 5.3.0 on d8dabb7 Platform: x64 sse4_1 Makefile:314: *** mixed implicit and normal rules: deprecated syntax Makefile:319: warning: overriding recipe for target 'obj-dreamcast-x64' Makefile:315: warning: ignoring old recipe for target 'obj-dreamcast-x64' Makefile:318: *** mixed implicit and normal rules: deprecated syntax Makefile:319: https://github.com/reicast/reicast-emulator/issues/883 warning: overriding recipe for target 'sse4_1/%.build_obj' Makefile:315: warning: ignoring old recipe for target 'sse4_1/%.build_obj' Makefile:323: warning: overriding recipe for target 'obj-dreamcast-x64' Makefile:319: warning: ignoring old recipe for target 'obj-dreamcast-x64' Makefile:322: *** mixed implicit and normal rules: deprecated syntax Makefile:323: warning: overriding recipe for target 'sse4_1/%.build_obj' Makefile:319: warning: ignoring old recipe for target 'sse4_1/%.build_obj' make: *** No rule to make target '../../core/%.S', needed by 'obj-dreamcast-x64'. Stop. make: Leaving directory '/tmp/makepkg/reicast-emulator-git/src/reicast-emulator/shell/linux' greetings Marc34 commented Dec 16, 2015 Same problem on Linux Mint. Reicast - A Dreamcast Emulator member Holzhaus commented Dec 16, 2015 Please try again with latest master. sl1pkn07 commented Dec 16, 2015 same Reicast - A Dreamcast Emulator member Holzhaus commented Dec 17, 2015 What version of GNU Make are you using? EDIT: Nevermind, i think i found the issue. Holzhaus added a commit that closed this issue Dec 17, 2015 Holzhaus http://eli.thegreenplace.net/2011/12/15/understanding-lvalues-and-rvalues-in-c-and-c run into these terms are in compiler error & warning messages. For example, compiling the following with gcc: int foo() {return 2;} int main() { foo() = 2; return 0; } You get: test.c: In function 'main': test.c:8:5: error: lvalue required as left operand of assignment True, this code is somewhat perverse and not something you'd write, in c but the error message mentions lvalue, which is not a term one usually finds in C/C++ tutorials. Another example is compiling this code with g++: int& foo() { return 2; } Now the error is: testcpp.cpp: In function 'int& foo()': testcpp.cpp:5:12: error: invalid initialization of non-const reference of type 'int&' from an rvalue of type 'int' Here again, lvalue and rvalue the error mentions some mysterious rvalue. So what do lvalue and rvalue mean in C and C++? This is what I intend to explore in this article. A simple definition This section presents an intentionally simplified definition of lvalues and rvalues. The rest of the article will elaborate on this definition. An lvalue (locator value) represents an object that occupies some identifiable location in memory (i.e. has an address). rvalues are defined by exclusion, by saying that every expression is either an lvalue or an rvalue. Therefore, from the above definition of lvalue, an rvalue is an expression that does not represent an object occupying some identifiable location in memory. Basic examples The terms as defined above may appear vague, which is why it's important to see some simple examples right away. Let's assume we have an integer variable defined and assigned to: int var; var = 4; An assignment expects an lvalue as its left operand, and var is an lvalue, because it is an object with an identifiable mem