https://stackoverflow.com › questions › 62353625

You can solve this by changing the reference from one that is not const -qualified to one that is. B operator+(const B& b) This version of the operator takes a const reference, which can bind to a temporary object. (Again, this is simply a rule of the language.

https://stackoverflow.com › questions › 20344670

I'm trying to compile my main.cpp, but I keep getting this error for two hours now. The issue here is to pass a function as a parameter but I think I'm doing something wrong. The compiler says it cannot find the function, but I included "newt_rhap(params)" in the "functions.h" already.

https://stackoverflow.com › questions › 11313600

Digging in it (version: 4.7.1), I've found what appear to be the relevant function in gcc/cp/pt.c: void print_candidates(tree fns) { const char *str = NULL; print_candidates_1 (fns, false, &str); gcc_assert (str == NULL); }

https://en.cppreference.com › w › cpp › language › overload_resolution

Candidate functions. Before overload resolution begins, the functions selected by name lookup and template argument deduction are combined to form the set of candidate functions. The exact details depend on the context in which overload resolution will take place.

http://guillaume.belz.free.fr › doku.php

Il indique à la ligne suivante qu'il connait une fonction f qui pourrait être candidate (“candidate function”), mais qui prend zéro argument (“requires 0 arguments”), alors que l'appel utilise un argument (“but 1 was provided”). Pour l'appel de la fonction g, le message est le suivant :

https://www.oreilly.com › library › view › c-primer-fifth › 9780133053043 › ch06lev2sec27.html

The first step of function matching identifies the set of overloaded functions considered for the call. The functions in this set are the candidate functions. A candidate function is a function with the same name as the called function and for which a declaration is visible at the point of the call.

https://cplusplus.com › forum › beginner › 274491

The issue is in how you call the function, not how you declare it. > requires 2 arguments, but 0 were provided. You tried to call makecube(); You should have, for example, if you're calling directly from main makecube(argc,argv);

https://clang.llvm.org › diagnostics.html

Clang captures and accurately tracks range information for expressions, statements, and other constructs in your program and uses this to make diagnostics highlight related information.

https://github.com › llvm › llvm-project › issues › 56394

When a template field is initialized via designated initializers, clang fails to deduce the final specialization. Code example: template < typename T> structFoo { int foo; T bar; }; intmain () { constexprauto foo = Foo { . foo = 1, . bar = + [] () { return5; } }; constexprauto eval = foo. bar (); return0; }

https://cplusplus.com › forum › beginner › 204305

int main() int arr[2][3]; fnc( arr ); fnc( arr ); ^~~. 'int [2][3]' to 'int **' for 1st argument. void fnc( int ** arr) { } You will need to cast your function call to match your function definition. But I thought, an array will be passed as pointer.