类C ++的Iota增量向量 [英] iota increment vector of a class c++

问题描述

我最近一直在利用< numeric> iota 语句来增加类型 int 。但是现在我试图使用该语句来增加具有2个成员的显式类。

I recently have been taking advantage of the <numeric> iota statement for incrementing a vector of type int. But now I am trying to use the statement for incrementing an explicit class with 2 members.

所以这是整数向量的用法：

So here is the usage with a vector of integers:

vector<int> n(6);
iota(n.begin(), n.end(), 1);

鉴于 Obj 类具有整数成员称为 m 。构造函数将 m 初始化为其相应的整数参数。这是我现在要执行的操作：

Given that Obj class has an integer member called m. The constructor initializes m to its corresponding integer argument. Here is what I am trying to do now:

vector<Obj> o(6);
iota(o.begin(), o.end(), {m(1)});

我试图使类增量重载有点像这样：

I've attempted making a class increment overload somewhat like this:

Obj& operator ++() {
    *this.m++;
    return *this;
}

但是我认为我的构造函数不是为这种重载而设计的，反之亦然。如何修改构造函数和重载以使用iota递增对象成员？

But I think either my constructor is not designed for this overload or vice versa. How can I modify my constructor and overload to increment an object member with iota? Thanks in advance!

推荐答案

我不确定我是否理解您的问题。以下代码是否符合您的要求？

I am not sure I understand your question. Does the following code match what you want?

#include <algorithm>
#include <iostream>
#include <vector>

class Object {
 public:
  Object(int value = 0)
      : m_value(value) { }
  Object& operator++() {
    m_value++;
    return *this;
  }
  int value() const {
    return m_value;
  }
 private:
  int m_value;
};

int main() {
  std::vector<Object> os(10);
  std::iota(os.begin(), os.end(), 0);
  for(const auto & o : os) {
    std::cout << o.value() << std::endl;
  }
}

在OS X 10.7.4上使用gcc 4.8编译得到：

Compiled with gcc 4.8 on OS X 10.7.4 I get:

$ g++ iota-custom.cpp -std=c++11
$ ./a.out 
0
1
2
3
4
5
6
7
8
9

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