逻辑NSexpression评估



此代码有效:

NSString* equation = @"2.5*3";
NSExpression* expresion = [NSExpression expressionWithFormat:equation, nil];
NSNumber* result = [expresion expressionValueWithObject:nil context:nil];
NSLog(@"%@", result); // 7.5

但这不是,它最终以NSInvalidArgumentException

NSString* equation = @"2.5<=3";
NSExpression* expresion = [NSExpression expressionWithFormat:equation, nil];
NSNumber* result = [expresion expressionValueWithObject:nil context:nil];
NSLog(@"%@", result); //I wanted result to be 1, as the expression is true

有人知道是否有一种使用nsexpression的方法来评估这样的逻辑表达式?

谢谢。

使用nspredicate代替nsexpression:

NSString* equation = @"2.5<=3";
NSPredicate* pre = [NSPredicate predicateWithFormat:equation];
BOOL result = [pre evaluateWithObject:nil];
NSLog(@"%@", (result ? @"YES" : @"NO"));

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