pass-by-reference

问题 Here is a simple example; template <typename T> void foo(T t) {} std::string str("some huge text"); foo(str); My question is how can I force the compiler to pass str by reference without modifying function foo? 回答1: Pass the reference type explicitly: template <typename T> void foo(T t) {} int main() { std::string str("some huge text"); foo<std::string&>(str); } This does modify the function instantiation that you get (by generating a void foo<std::string&>(std::string& t) ), but it doesn't

问题 To cut a long story short I have the following function as part of my framework: public function use_parameters() { $parameters = func_get_args(); $stack = debug_backtrace(); foreach($stack[0]['args'] as $key => &$parameter) { $parameter = array_shift($this->parameter_values); } } Where $this->parameter_values = array('value1', 'value2', 'value3', 'value4', 'value5', ...); Which is used in the following context: $instance->use_parameters(&$foo, &$bah); Assigning: $foo = 'value1'; $bah =

问题 To cut a long story short I have the following function as part of my framework: public function use_parameters() { $parameters = func_get_args(); $stack = debug_backtrace(); foreach($stack[0]['args'] as $key => &$parameter) { $parameter = array_shift($this->parameter_values); } } Where $this->parameter_values = array('value1', 'value2', 'value3', 'value4', 'value5', ...); Which is used in the following context: $instance->use_parameters(&$foo, &$bah); Assigning: $foo = 'value1'; $bah =

问题 To cut a long story short I have the following function as part of my framework: public function use_parameters() { $parameters = func_get_args(); $stack = debug_backtrace(); foreach($stack[0]['args'] as $key => &$parameter) { $parameter = array_shift($this->parameter_values); } } Where $this->parameter_values = array('value1', 'value2', 'value3', 'value4', 'value5', ...); Which is used in the following context: $instance->use_parameters(&$foo, &$bah); Assigning: $foo = 'value1'; $bah =

问题 I think it's illegal to pass a reference to a reference in C++.However ,when I run this code it gives me no error. void g(int& y) { std::cout << y; y++; } void f(int& x) { g(x); } int main() { int a = 34; f(a); return 0; } Doesn't the formal parameter of g() qualify as a reference to a reference ?? 回答1: 1) There is nothing wrong with passing a reference to a reference (it is what the move-constructor and move-assignment operators use - though it is actually called a rvalue-reference). 2) What

问题 In .NET, strings are immutable and are reference type variables. This often comes as a surprise to newer .NET developers who may mistake them for value type objects due to their behavior. However, other than the practice of using StringBuilder for long concatenation esp. in loops, is there any reason in practice that one needs to know this distinction? What real-world scenarios are helped or avoided by understanding the value-reference distinction with regard to .NET strings vs. just

问题 In .NET, strings are immutable and are reference type variables. This often comes as a surprise to newer .NET developers who may mistake them for value type objects due to their behavior. However, other than the practice of using StringBuilder for long concatenation esp. in loops, is there any reason in practice that one needs to know this distinction? What real-world scenarios are helped or avoided by understanding the value-reference distinction with regard to .NET strings vs. just

问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: What does T&& mean in C++11? For some reason, this is eluding my intuition, and I cannot find any explanation on the internet. What does it mean for a C++ function to take a reference of a reference? For example: void myFunction(int&& val); //what does this mean?! I understand the idea of passing-by-reference, so void addTwo(int& a) { a += 2; } int main() { int x = 5; addTwo(x); return 0; } works and is

问题 I made a simple program in c++ to compare performance between two approaches - pass by value and pass by reference. Actually pass by value performed better than pass by reference. The conclusion should be that passing by value require fewer clock-cycles (instructions) I would be really glad if someone could explain in detail why pass by value require fewer clock-cycles. #include <iostream> #include <stdlib.h> #include <time.h> using namespace std; void function(int *ptr); void function2(int

问题 If I have a function in program int main(){ char *name = "New Holland"; modify(name); printf("%s\n",name); } that calls this function void modify(char *s){ char new_name[10] = "Australia"; s = new_name; /* How do I correct this? */ } how can I update the value of the string literal name (which now equals new Holland) with Australia. The problem I think that I face is the new_name is local storage, so after the function returns, the variable is not stored 回答1: Try this: #include <stdio.h> void