C ++ vector: Call of erase and non-argument Constructor (Code tutorial), vectorerase

C ++ vector: Call of erase and non-argument Constructor (Code tutorial), vectorerase

Vector: erase

Delete the element of C ++ vector. The source code is as follows:

template <class _Tp, class _Allocator>inline _LIBCPP_INLINE_VISIBILITYtypename vector<_Tp, _Allocator>::iteratorvector<_Tp, _Allocator>::erase(const_iterator __position){#if _LIBCPP_DEBUG_LEVEL >= 2    _LIBCPP_ASSERT(__get_const_db()->__find_c_from_i(&__position) == this,        "vector::erase(iterator) called with an iterator not"        " referring to this vector");#endif    _LIBCPP_ASSERT(__position != end(),        "vector::erase(iterator) called with a non-dereferenceable iterator");    difference_type __ps = __position - cbegin();    pointer __p = this->__begin_ + __ps;    this->__destruct_at_end(_VSTD::move(__p + 1, this->__end_, __p));    this->__invalidate_iterators_past(__p-1);    iterator __r = __make_iter(__p);    return __r;}

C ++ moves all the elements behind the deleted part forward as a set.

Therefore, assume that we need to clear the elements, and do not write as follows:

int main(){    vector<int> v_int;    for(int i=0;i<10;i++){        v_int.push_back(i+1);    }    int size = v_int.size();    vector<int>::iterator it = v_int.begin();    while(size--) {        cout<<*it<<" ";        v_int.erase(it++);   // attention !        cout<<"size: "<<v_int.size()<<endl;    }    return 0;}

The result is as follows:

1 size: 93 size: 85 size: 77 size: 69 size: 5

In this example, you can change it ++ to it to clear all the elements.

No-argument constructor call

Will Base instance () call the class's no-argument constructor?

The answer is no. C ++ interprets it as a function that returns the Base object.

#include <iostream>#include <vector>#include <string>using namespace std;class Base{public:    Base(){ cout<<"Base()..\n"; }    Base(string str){ cout<<str<<endl; }    ~Base(){ cout<<"~Base().. \n"; }};int main(){    //Base ins(); //empty parentheses interpreted as a function declaration    Base ins;     //ok    Base ins2("hello world");  //ok    return 0;}

However, it is interesting that the constructor with parameters can be called in this way.

Sometimes, we encounter such a function:

void show(const Base &b){    //...}

We need to pass a variable stored in the stack to this function. To construct such a variable, we can simply write Base () in this way. It is equivalent to Base B.

That is:

    //show(Base("hello"));  //ok. output: hello     it's equal to 'Base b("hello"); show(b);'    //show(Base());         //ok. output: Base()..  it's equal to 'Base b; show(b);'