无需发送任何内容即可返回函数(Get return of a function without sending anything)
我有一个函数头:
double countThis(double counter);然后,在我的主要内容中我这样做:
double test = 10; countThis(test);然后是功能:
double countThis(double counter) { double counted = counter; return counted; }在底部,我有一个最后一个函数,在这里我想得到double counted而不必做countThis(something) ,我只想从上一次调用返回在main中获得并获得值10(计数) )。
I have a function header:
double countThis(double counter);Then, in my main I do this:
double test = 10; countThis(test);Then comes the function:
double countThis(double counter) { double counted = counter; return counted; }On the bottom, I have one last function, and here I want to get double counted without having to do countThis(something), I just want to get the return from the previous call that was made in main and get the value 10 (counted).
最满意答案
实现此类持久性的一种方法是使用类并定义该类的实例 :
struct Counter { double counted; double countThis(double counter) { return counted = counter; // assign counter to counted, and return that value. } };在使用时:
int main() { Counter c; c.countThis(10); // c.counted contains the last value sent to countThis, in this case, 10 }实例 c用于保存您传递给countThis的值。
One way to achieve this sort of persistence is to use a class and define an instance of that class:
struct Counter { double counted; double countThis(double counter) { return counted = counter; // assign counter to counted, and return that value. } };At the point of use:
int main() { Counter c; c.countThis(10); // c.counted contains the last value sent to countThis, in this case, 10 }The instance c is used to persist the value that you pass to countThis.
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