stl::带原始指针的迭代器 [英] stl::iterators with raw pointers

问题描述

我想对 C++ 数组使用迭代器，但也要使用原始指针.我可以使用静态向量:

I want to use iterators with C++ arrays, but with raw pointers too. I can do with a static vector:

#define SIZE 10
int vect[SIZE] = {0};
vect[3] = 5;
int* p = std::find(std::begin(vect), std::end(vect), 5);
bool success = p != std::end(vect);

如何使用原始指针(可能是堆分配的向量)来做到这一点?当然编译器不知道数据的大小，所以这段代码

How can be possible to do it with a raw pointer (maybe a heap allocated vector)? Of course the compiler does not know the size of the data, so this code

int* pStart = vect;
std::find(std::begin(pStart), std::end(pStart), 5);

给予

error C2784: '_Ty *std::begin(_Ty (&)[_Size])' : 
could not deduce template argument for '_Ty (&)[_Size]' from 'int *'

是否可以让 begin() 和 end() 意识到它?

Is it possible to make begin() and end() aware of it?

推荐答案

不，不能在指针上使用 std::begin 和 std::end.不像数组的大小是类型的一部分，因此可以推导出指针不保存它指向的东西的大小.在您使用指针的情况下，您必须使用

No it is not possible to use std::begin and std::end on a pointer. Unlike an array whose size is part of the type and therefor deducible a pointer does not hold the size of the thing it points to. In your case with a pointer you would have to use

std::find(pStart, pStart + SIZE, 5);

避免这种情况的方法是使用 std::vector 当您在编译时不知道 szie 将是什么时.它将为您管理内存并提供begin 和end 成员函数.

The way to avoid this though is to use std::vector when you are not going to know what the szie will be at compile time. It will manage the memory for you and provides begin and end member functions.

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