Since the questioner knows that observing the assembly window, it is easy to find that the copy constructor of the object is called there, so here is by value
In fact, all objects in C++ are treated as by value. If you need by reference, you need to change the function signature to const string& obj
1. If the object is passed into the function as a value parameter, such as func (const string obj), the object will be copied and a temporary object will be generated, which affects performance; 2. If the object is used as a reference or pointer When parameters are passed in, such as func (const string &obj), fun (const string *obj), the object will not be copied or copied, and no temporary object will be generated.
Because the way you pass parameters is by value, the copy constructor is called, which can be replaced by string &x; this is passing a reference. In addition, c++11 has newly added the ability to use string && x to pass temporary variables
The method of passing parameters in C++ is value copy (by value) or reference copy (by reference), depending on how you name the formal parameters.
Example 1: Pass by value.
/* function declaration of transferring argument by value */
void func(string s) {
// operates on string object s
// note that here s is a copy of the string object you transferred in so any function-scope change will not affect invoker environment.
}
/* example usage of aforementioned function */
string str;
func(str);
// after func call, str will keep the same
Example 2: Pass by reference.
/* function declaration */
void func(string& s) {
// operate on string object s
// local change will be applied to the argument you transferred in
}
/* example usage */
string str;
func(str);
// at this time, str will be changed with the modification func() made to it
Example 3: Pass by constant reference (in actual work, this calling method is the most commonly used, it can effectively reduce the memory and CPU time consumption of string copy) .
/* function declaration */
void func(const string& s) {
// transfer in a read-only reference of string object s
// no copy will be made against s, neither any change allowed upon it
}
/* example usage */
string str;
func(str);
// str is not supposed to be changed after func() invoke.
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Yes.
Not necessarily, it depends on the definition of the copy/move constructor. For example: Value-like class and Pointer-like class.
Since the questioner knows that observing the assembly window, it is easy to find that the copy constructor of the object is called there, so here is by value
In fact, all objects in C++ are treated as by value. If you need by reference, you need to change the function signature to const string& obj
1. If the object is passed into the function as a value parameter, such as func (const string obj), the object will be copied and a temporary object will be generated, which affects performance;
2. If the object is used as a reference or pointer When parameters are passed in, such as func (const string &obj), fun (const string *obj), the object will not be copied or copied, and no temporary object will be generated.
Because the way you pass parameters is by value, the copy constructor is called, which can be replaced by string &x; this is passing a reference. In addition, c++11 has newly added the ability to use string && x to pass temporary variables
The method of passing parameters in C++ is value copy (by value) or reference copy (by reference), depending on how you name the formal parameters.
Example 1: Pass by value.
Example 2: Pass by reference.
Example 3: Pass by constant reference (in actual work, this calling method is the most commonly used, it can effectively reduce the memory and CPU time consumption of string copy) .