问题描述
我想验证用户创建的表达式(如2 + 2,5 + 7或更复杂)。我使用NSExpression类来解析和计算这个表达式。这是我的Playground代码:
I want to validate user created expressions (like "2+2", "5+7" or more complex). I use NSExpression class to parse and calculate this expressions. This is my Playground code:
import UIKit
let string = "2+2"
var ex:NSExpression?
do {
ex = NSExpression(format: string)
}
catch {
print("String is not valid expression")
}
if let result = ex?.expressionValue(with: nil, context: nil) as! NSNumber? {
print("result is \(result)")
}
当我使用有效的表达式(2 + 2) - 我得到结果。但有时用户可以提供错误的字符串(例如2+)。有了这个字符串,我的应用程序会崩溃:
When I use valid expression ("2+2") - I get the result. But sometime user can provide wrong string ("2+" for example). With this string my app crashes with this:
*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Unable to parse the format string "2+ == 1"'
我不明白我怎么可以捕获这个异常,为什么上面的代码不这样做。现在我使用Objective C类(具有相同的逻辑),从我的swift代码调用这个方法,在该类中我真的可以捕获这样的异常:
I don't understand how I can catch this exception and why code above don't do this. Now I use Objective C class (with same logic), calling this method from my swift code, and in that class I really can catch such exception:
+(NSNumber *)solveExpression:(NSString *)string
{
id value;
@try {
NSExpression *ex = [NSExpression expressionWithFormat:string];
value = [ex expressionValueWithObject:nil context:nil];
}
@catch (NSException *e) { }
return value;
}
这个工作,我可以得到正确的解析状态(nil表示表达式的问题)和结果(NSNumber),但我真的想了解如何在Swift中正确和完整地完成所有这些事情。
This works and I can get correct parse state (nil means problems with expression) and result (NSNumber), but I really want to understand how to do all this things correct and entirely in Swift.
推荐答案
是什么书籍使用Swift与Cocoa和Objective-C :
This is what the book Using Swift with Cocoa and Objective-C has to say:
[我的粗体]
刚刚删除了 NSExpression的引用
我看不出有关这个问题的简单方法。上面的引用建议写一些Objective-C代码来做。最简单的方法可能是创建一个C函数:
Having just skimmed the reference for NSExpression
I can't see a simple way around the issue. The quote above recommends writing a a bit of Objective-C code to do it. The simplest way is probably to create a C function:
声明:
extern NSExpression* _Nullable makeExpression(NSString* format _Nonnull);
定义
NSExpression* _Nullable makeExpression(NSString* format _Nonnull)
{
NSExpression* ret = nil;
@try
{
// create and assign the expression to ret
}
@catch(NSException* e)
{
// ignore
}
return ret;
}
该函数对于错误的表达式返回nil。
The function returns nil for expressions that are in error.
您可能会添加一个 NSError **
参数,以便在发生故障时使用。您也可以在 NSExpression
中的类别中使用此方法,然后返回无效的错误/填充NSError模式可能会导入Swift作为Swift投掷方法。
You could probably add an NSError**
parameter to be used when there is a failure. You could also probably make this a method in a category on NSExpression
and then the return nil for error/fill in NSError pattern would probably be imported to Swift as a Swift throwing method.
顺便说一下,Objective-C异常并没有真正保证让程序保持一致状态。在这种情况下,手指越过它就可以。
I should say, by the way, that an Objective-C exception is not really guaranteed to leave your program in a consistent state. Fingers crossed it is OK in this case.
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