为什么我可以在没有弦乐的情况下运行我的Getline代码？如何使用StringStream使此代码起作用？
#include<iostream>
#include<string>
using namespace std;
int main() {
string randomwords,temp;
getline(cin,randomwords);
while(getline(randomwords,temp,' ')) {
cout<<temp<<endl;
}
return 0;
}
#include<iostream>
#include<string>
using namespace std;
int main() {
string randomwords,temp;
getline(cin,randomwords);
while(getline(randomwords,temp,' ')) {
cout<<temp<<endl;
}
return 0;
}
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std :: getline的第一个参数是std :: basic_istream。std :: basic_string和std :: basic_istream之间没有转换std :: Basic_string）作为std :: getline的第一个参数。这是C ++的基本规则，函数的参数必须具有匹配类型，或具有几种可用于将一种类型对象转换为另一种类型的对象之一。这里没有，所以这就是为什么它不起作用的原因。但是，
std :: Basic_istringStream具有一个重载的构造函数，该构造函数将std :: Basic_string作为参数。通常，它可以用作隐式转换，但是该特定的构造函数是explicit构造函数，该构建器禁止将其用于隐式类型转换中。因此，您将自己完成工作：明确地从字符串构建输入流，std :: getline将很乐意使用它。任务完成了。std::getline's first parameter is astd::basic_istream. There is no conversion between astd::basic_stringand astd::basic_istream, so you cannot pass astd::string(a specialization ofstd::basic_string) as a first parameter tostd::getline. This is a fundamental rule of C++, parameters to functions must have matching types or have one of several conversions that can be used to convert an object of one type to the other one. There are none here, so that's why it won't work.However,
std::basic_istringstreamhas an overloaded constructor that takes astd::basic_stringas a parameter. Normally that can be used as an implicit conversion, but this particular constructor is anexplicitconstructor which prohibits it from being used in implicit type conversions. Therefore you'll just do the job yourself: construct an input stream from a string explicitly, andstd::getlinewill happily use it. Mission accomplished.