如何使用std :: copy来打印用户定义的类型 [英] How to use std::copy for printing a user defined type

问题描述

下面的代码非常适合打印std::string

Below is the code that works perfectly for print the values of type std::string

std::vector<std::string> v;
v.push_back("this");
v.push_back("is");
v.push_back("a");
v.push_back("test");
std::copy(v.begin(),v.end(),std::ostream_iterator<std::string>(std::cout,","));

但是当我尝试打印用户定义的类型(结构)时，代码无法编译:

But when I am trying to print a user defined type (a structure), code is not compiling:

struct Rec
{
    int name;
    int number;
    int result;
};
int main() 
{
    Rec rec1 = {1,1,1};
    Rec rec2 = {2,1,1};
    Rec rec3 = {3,1,1};
    Rec rec4 = {4,1,1};
    Rec rec5 = {4,1,1};

    std::vector<Rec> v;
    record.push_back(rec1);
    record.push_back(rec2);
    record.push_back(rec3);
    record.push_back(rec4);
    record.push_back(rec5);

    std::copy(v.begin(),v.end(),std::ostream_iterator<Rec>(std::cout,","));

    return 1;
}

我在这里想念什么?

推荐答案

来自 std :: ostream_iterator

std::ostream_iterator是写的单遍OutputIterator 类型T的连续对象进入std::basic_ostream对象，用于 它是使用operator<<构造的.可选的分隔符字符串 每次写操作后，都会将其写入输出流.写 迭代器(是否已取消引用)时执行操作 被分配给. std::ostream_iterator递增是空操作.

std::ostream_iterator is a single-pass OutputIterator that writes successive objects of type T into the std::basic_ostream object for which it was constructed, using operator<<. Optional delimiter string is written to the output stream after every write operation. The write operation is performed when the iterator (whether dereferenced or not) is assigned to. Incrementing the std::ostream_iterator is a no-op.

(经评论确认)

对于自定义记录，您不会超载operator <<.使用以下代码使运算符重载.

You are not overloading operator << for the custom record. Use the following code to overload the operator.

ostream& operator<<(ostream& os, const Rec& r)  
{  
    os << r.name << '-' << r.number << '-' << r.result;  
    return os;  
}

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