使用模板打印任何容器的所有数据 [英] use template to print all the data of any container
问题描述
我计划编写一个可以打印任何容器的所有数据的函数。换句话说,我可以使用不同的容器类型(例如vector,deque或list),也可以使用不同的数据类型(整数,双精度或字符串)来调用它。
模板函数可以传递编译器,但我不知道如何调用它。
I plan to write a function which can print all the data of any container. In other words, I can use with different containers type like vector, deque or list and that I can call it with different data types ( integers , double or string). The template function can pass the compiler, but I do not know how to call it.
#include <cstdlib>
#include <stdio.h>
#include <iostream>
#include <list>
using namespace std;
template <typename C, template <typename C> class M>
void print(M<C> data){
typename M<C>::iterator it;
for(it=data.template begin();it!=data.template end();++it){
cout<<*it<<" ";
}
cout<<endl;
}
int main(int argc, char** argv) {
list<int> data;
for(int i=0;i<4;i++){
data.push_back(i);
}
print<int>(data); //compile error
print<int, list>(data); //compile error
return 0;
}
错误消息:
main.cpp:35:20:错误:没有匹配函数可调用'print(std :: list&)'
main.cpp:35:20:注意:候选者是:
main.cpp:21:6:注意:模板类M> void print(M)
error message: main.cpp:35:20: error: no matching function for call to 'print(std::list&)' main.cpp:35:20: note: candidate is: main.cpp:21:6: note: template class M> void print(M)
一些相关线程:
用于多种容器和数据类型的模板函数
http://louisdx.github.io/cxx-prettyprint/
推荐答案
如注释中所述, std :: list
实际上具有多个模板参数(第二个参数是分配器)。此外,不需要 print
来获取两个模板参数;您可以简单地在整个容器类型上对其进行参数化:
As noted in the comments, std::list
actually has more than one template parameter (the second parameter is the allocator). Moreover, there is no need for print
to take two template parameters; you can simply parameterize it over the overall container type:
#include <iostream>
#include <iterator>
#include <list>
using namespace std;
template <typename C>
void print(const C &data){
for(auto it=begin(data);it!=end(data);++it){
cout<<*it<<" ";
}
cout<<endl;
}
int main(int argc, char** argv) {
list<int> data;
for(int i=0;i<4;i++){
data.push_back(i);
}
print(data);
return 0;
}
演示。
如另一答案所指出,如果您可以使用C ++ 11功能, range-for循环比上面的显式迭代器更好。如果不能(这意味着没有 auto
或 std :: begin
或 std: :end
),然后:
As pointed out in the other answer, if you can use C++11 features, a range-for loop is better than the explicit iterator use above. If you can't (which means no auto
or std::begin
or std::end
either), then:
template <typename C>
void print(const C &data){
for(typename C::const_iterator it=data.begin();it!= data.end();++it){
cout<<*it<<" ";
}
cout<<endl;
}
请注意,因为我们采用数据
通过const引用,我们需要使用 const_iterator
。
Note that since we take data
by const reference, we need to use const_iterator
.
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