从NSExpression捕获NSInvalidArgumentException [英] Catching NSInvalidArgumentException from NSExpression

问题描述

在我的代码中，我将字符串作为数学表达式进行评估，例如：

In my code I am evaluating strings as mathematical expressions for example:

NSString *formula=@"9*7";
NSExpression *expr =[NSExpression expressionWithFormat:formula];
NSLog(@"%@", [[expr expressionValueWithObject:nil context:nil]intValue]);

以上工作正常，但我将处理用户的动态输入，所以我需要能够捕获当用户输入错误数据时的异常，因此我需要能够在以下情况下捕获异常：

The above works fine but I will be handling dynamic input from users so I need to be able to catch the exception when the user enters faulty data, thus I need to be able to catch the exception in situations like the following:

NSString *formula=@"9*"; //note the deliberately invalid expression
NSExpression *expr =[NSExpression expressionWithFormat:formula];
@try {        
    [[expr expressionValueWithObject:nil context:nil]intValue];
}
@catch (NSException *exception) {
    NSLog(@"Exception");
}
@finally {
    NSLog(@"Finally");
}

然而，当我运行此代码时，我得到：

However when I run this code I get:

由于未捕获的异常'NSInvalidArgumentException'而终止应用程序，原因：'无法解析格式字符串9 * == 1'

Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Unable to parse the format string "9* == 1"'

有没有办法捕获此异常？或者是否有某种方法来测试表达式在我传递之前是否有效？

Is there some way to catch this exception? Or alternatively is there some way to test if an expression is valid before I pass it off?

谢谢！

推荐答案

此异常未被当前代码捕获的原因是该行引发了异常：

The reason this exception is not caught with your current code is that the exception is being thrown from this line:

NSExpression *expr =[NSExpression expressionWithFormat:formula];

您需要将此行移至 @try 阻止。

You need to move this line into the @try block.

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