将float变量作为参数传递 [英] passing float variable as parameter

问题描述

我正在尝试编写一个带有float参数的方法，并使用performselector调用它，但是这样做却出错了.以下是我的代码:

I am trying to write a method with float parameter and call it using performselector but i am getting error in doing this. Following is my code:

[发送器performSelector:带有对象的选择器:进度/总计];

[sender performSelector:selector withObject:progress/total];

这里的进度和总数都是浮动变量.

here progress and total both are float variabal.

我正在尝试在其他类中调用以下方法

I am trying to call following method in different class

-(void) updateProgress:(float)fl {

}

推荐答案

您需要传递一个真实的对象，而不是int或float这类基本类型之一.

You need to pass a real object, not one of the basic types like int or float.

将其包装为NSNumber对象:

[sender performSelector:selector withObject:[NSNumber numberWithFloat:progress/total]];

-(void) updateProgress:(NSNumber *)aProgress {
   float fProgress = [aProgress floatValue];
}

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