C++ 对列表

C++ List of Pairs

提问by Vlad from Moscow

While getting comfortable with the C++ STL I ran into this issue using a list of pair objects.

在熟悉 C++ STL 的同时，我使用配对对象列表遇到了这个问题。

int count (std::string input, std::vector<std::string> screen){

  std::string::iterator it = input.begin();

  std::pair<std::string, int> freq[10];
  std::list<std::pair<std::string,int>  > freq2;

  freq2.push_back(std::make_pair(*it,0)); //ERR:No instance of fn matches arg list 
  freq2.push_back(std::make_pair('S',0)); //ERR:No instance of fn matches arg list
  freq2.push_back(std::make_pair("S",0)); //This is fine

  freq[0] = std::make_pair(*it,0);        //This is fine
  freq[0] = std::make_pair("S",0);        //This is fine
  freq[0] = std::make_pair('S',0);        //This is fine

  return 1;

}

Both freq and freq2 are quite similar except freq2 is just a list. Freq is able to accept chars, strings, and the iterator pointer and both pairs (freq,freq2) are declared as pairs. Not sure why this is happening, any tips? Thanks.

freq 和 freq2 非常相似，除了 freq2 只是一个列表。Freq 能够接受字符、字符串和迭代器指针，并且两个对 (freq,freq2) 都被声明为对。不知道为什么会这样，有什么提示吗？谢谢。

edit: It makes a lot more sense now, all the responses I got were really helpful, thank you guys!

编辑：现在更有意义了，我得到的所有回复都非常有帮助，谢谢你们！

回答by Slava

There is no implicit constructor in std::stringthat accepts charand can be used to convert type charto std::string. What you can use is this constructor:

没有隐式构造函数std::string接受char并且可用于将类型转换char为std::string. 你可以使用的是这个构造函数：

basic_string( size_type count, CharT ch, const Allocator& alloc = Allocator() );

basic_string( size_type count, CharT ch, const Allocator& alloc = Allocator() );

so

所以

freq2.push_back(std::make_pair( std::string( 1, *it ),0));

and so on

等等

回答by Vlad from Moscow

Class std::basic_stringhas no constructor that has one parameter of type char.So these statemenets are invalid

类std::basic_string没有具有一个类型参数的构造函数char。所以这些语句是无效的

  freq2.push_back(std::make_pair(*it,0)); //ERR:No instance of fn matches arg list 
  freq2.push_back(std::make_pair('S',0)); //ERR:No instance of fn matches arg list

Speaking more precisely in these statements there is an attempt to build an object of type std::pair<std::string, int>from an object of type std::pair<char, int>. This requires that there would be a constructor of class std::stringthat has parameter of type char. However there is no such a constructor.

更准确地说，在这些语句中，尝试std::pair<std::string, int>从type的对象构建 type的对象std::pair<char, int>。这要求将有一个std::string具有类型参数的类的构造函数char。但是没有这样的构造函数。

As for these statements

至于这些说法

  freq[0] = std::make_pair(*it,0);        //This is fine
  freq[0] = std::make_pair('S',0);        //This is fine

then there is used an assignment operator of class std::basic_string. The class has an overloaded assignment operator that accepts an object of type char.

然后使用了 class 的赋值运算符std::basic_string。该类有一个重载的赋值运算符，它接受类型为 的对象char。

basic_string& operator=(charT c);

So these statements are correct.

所以这些说法是正确的。

Consider the following example

考虑下面的例子

#include <iostream>
#include <string>
#include <utility>

int main() 
{
    std::pair<std::string, int> p1 = { "a", 1 };
    std::pair<char, int> p2 = { 'B', 2 };

    p1 = p2;

    std::cout << p1.first << '\t' << p1.second << std::endl;

    return 0;
}

The output is

输出是

B   2

As class std::pairhas a template assignment operator and class std::stringin turn has assignment operator that has parameter of type char then this code is compiled successfully.

由于类std::pair具有模板赋值运算符，而类std::string又具有具有 char 类型参数的赋值运算符，因此此代码编译成功。

If you would write for example

如果你会写例如

std::pair<std::string, int> p1 = std::make_pair( 'a', 1 );

instead of

代替

    std::pair<std::string, int> p1 = { "a", 1 };

you would get an error because there is no conversion from std::pair<char, int>to std::pair<std::string, int>because there is no constructor from charto std::string.

你会得到一个错误，因为没有从std::pair<char, int>to 的转换，std::pair<std::string, int>因为没有构造函数 from charto std::string。

回答by Ian T. Small

You seem to be mixing char and std::string types. String iterators point to individual char values, not a std::string. Review http://www.cplusplus.com/reference/string/string/begin/for details.

您似乎混合了 char 和 std::string 类型。字符串迭代器指向单个字符值，而不是 std::string。查看http://www.cplusplus.com/reference/string/string/begin/了解详细信息。

Also, 'S'is NOT equivilent to "S"as the first is of type char while the second is a of type char* which points to an address in memory which contains a 'S'followed by a '\0'(the "null character" which is NOT the same as null).

此外，'S'不等于"S"第一个是 char 类型，而第二个是 char* 类型，它指向内存中的地址，其中包含'S'后跟 a '\0'（与 null 不同的“空字符”）。

freq2.push_back(std::make_pair(*it,0)); //The iterator is pointing to a literal character value, and not a string
freq2.push_back(std::make_pair('S',0)); //This is a literal character, and not a string literal
freq2.push_back(std::make_pair("S",0)); //This is fine and what you intended

The assignments to the array do so via an implicit conversion. See http://www.cplusplus.com/reference/utility/make_pair/.

对数组的赋值是通过隐式转换完成的。请参阅http://www.cplusplus.com/reference/utility/make_pair/。

freq[0] = std::make_pair(*it,0);        //Implicit conversion from Pair<char, int>
freq[0] = std::make_pair("S",0);        //Implicit conversion from Pair<char*, int>
freq[0] = std::make_pair('S',0);        //Implicit conversion from Pair<char, int>

A potentialwork around if you actually needthis conversion functionality would be to create a variable of type std::pair<std::string, int>and then pushing the value onto the list. I would only recommend this in a desperate and odd case though, and instead correct the intent of the code directly.

如果您确实需要此转换功能，一个潜在的解决方法是创建一个类型的变量，然后将值推送到列表中。我只会在绝望和奇怪的情况下推荐这个，而是直接更正代码的意图。std::pair<std::string, int>

回答by PaulMcKenzie

The issue is that you're attempting to construct a std::stringfrom a single char. There is no such constructor for std::string.

问题是您试图std::string从单个字符构造 a 。没有这样的构造函数std::string。