[英]Running function with std::istream
I am trying to input a value in my function, which looks like this: 我正在尝试在函数中输入一个值,如下所示:
int funkcija( std::istream & in ) {
int value(0);
in >> value;
if(not in) throw std::exception();
if( value%2 == 0 ) {
return (value/2);
}
else return (value*3)+1;
}
When I try to run it: 当我尝试运行它时:
int i(0);
std::cout << "Input a number: ";
std::cin >> i;
funkcija(i);
I get an error: ..\\working.cpp:17:14: error: invalid initialization of reference of type 'std::istream& {aka std::basic_istream&}' from expression of type 'int' ..\\working.cpp:7:5: error: in passing argument 1 of 'int funkcija(std::istream&)' 我收到一个错误:.. \\ working.cpp:17:14:错误:类型'int'.. \\ working.cpp类型的表达式对'std :: istream&{aka std :: basic_istream&}'引用的初始化无效:7:5:错误:传递'int funkcija(std :: istream&)'的参数1
What does it mean and how to solve it? 这是什么意思,以及如何解决? Thank you!
谢谢!
You are trying to pass the integer you've already read, try: 您正在尝试传递已经读取的整数,请尝试:
std::cout << "Input a number: ";
int i = funkcija(std::cin);
std::cout << i << " ";
While this would work it seems strange. 尽管这行得通,但似乎很奇怪。 Consider separating the input- and output-handling from the calculation to improve your design.
考虑将输入和输出处理与计算分开以改善设计。 Change the function to:
将该功能更改为:
int funkcija( int value ) {
if( value%2 == 0 ) {
return (value/2);
}
else return (value*3)+1;
}
and maybe call it like this: 也许这样称呼它:
std::cout << "Input a number: ";
int i;
if( !( std::cin >> i ) ) throw std::exception();
do {
i = funkcija( i );
std::cout << i << " ";
} while( i != 1 );
i is of type int not istreeam - you're passing i to the function, therefore its complaining that it's an int and not an istream. 我的类型是int而不是istreeam-您正在将i传递给函数,因此它抱怨它是int而不是istream。 You can probably pass in std::cin directly to the function.
您可能可以将std :: cin直接传递给该函数。
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