如何从NSExpression的expressionValueWithObject：上下文方法获得浮点数？ [英] How to get floating point number from the NSExpression's expressionValueWithObject:context method?
问题描述
我已经实现了一个自定义计算器，我正在使用下面的代码来评估类似5 + 3 * 5-3的算术表达式。
< （NSNumber *）evaluateArithmeticStringExpression：（NSString *）表达式{
NSNumber * calculatedResult = nil;
@try {
NSPredicate * parsed = [NSPredicate predicateWithFormat：[expression stringByAppendingString：@= 0]];
NSExpression * left = [（NSComparisonPredicate *）parsed leftExpression];
calculatedResult = [left expressionValueWithObject：nil context：nil];
@catch（NSException * exception）{
NSLog（@Input is not a expression ...！）;
}
@finally {
return calculatedResult;
$ b 但是当我使用除法运算的整数时结果只得到整数。让我们说5/2我得到2结果。由于整数除法对编程的撼动是正确的。
$ b
但是我需要浮点结果。
<我怎样才能得到它，而不是扫描表达式字符串，并将整数除数替换为浮点数。在我们的例子5 / 2.0或5.0 / 2中。
解决方案我自己找到了。 （NSNumber *）evaluateArithmeticStringExpression：（NSString *）表达式{
NSNumber * calculatedResult = nil;
pre $
@try {
NSPredicate * parsed = [NSPredicate predicateWithFormat：[NSString stringWithFormat：@1.0 *％@ = 0，expression]];
NSExpression * left = [（NSComparisonPredicate *）parsed leftExpression];
calculatedResult = [left expressionValueWithObject：nil context：nil];
@catch（NSException * exception）{
NSLog（@Input is not a expression ...！）;
}
@finally {
return calculatedResult;
$ b $ p
$ b $ p <
NSPredicate * parsed = [NSPredicate predicateWithFormat：[NSString stringWithFormat：@1.0 * ％@ = 0，expression]];
注意：感谢@Martin R，但是我的问题不是关于整数除法，而是完全关于NSExpression 。我已经明确地排除了我的最后一句话。
@Zaph，这里使用异常处理是有很强的理由的。这是一个地方，我的方法是接受用户输入，我们的用户可以输入类似w * g和 - expressionValueWithObject：context：将抛出一个异常，我必须避免我的应用程序异常终止。如果用户输入了一个有效的表达式，那么他/她将以NSNumber的形式得到一个答案，否则就是一个零的NSNumber对象。
I have already implemented a custom calculator where I am using following code to evaluate the arithmetic expression something like 5+3*5-3.
- (NSNumber *)evaluateArithmeticStringExpression:(NSString *)expression {
NSNumber *calculatedResult = nil;
@try {
NSPredicate * parsed = [NSPredicate predicateWithFormat:[expression stringByAppendingString:@" = 0"]];
NSExpression * left = [(NSComparisonPredicate *)parsed leftExpression];
calculatedResult = [left expressionValueWithObject:nil context:nil];
}
@catch (NSException *exception) {
NSLog(@"Input is not an expression...!");
}
@finally {
return calculatedResult;
}
}
But when I am having integers with division operation I am getting only integer as a result. Let's say for 5/2 I am getting 2 as a result. It's right for the shake of programming because of the integer division.
But I need floating point result.
How can I get it rather scanning the expression string and replace the integer divisor as a floating point. In our example 5/2.0 or 5.0/2.
解决方案 I found it by myself.
- (NSNumber *)evaluateArithmeticStringExpression:(NSString *)expression {
NSNumber *calculatedResult = nil;
@try {
NSPredicate * parsed = [NSPredicate predicateWithFormat:[NSString stringWithFormat:@"1.0 * %@ = 0", expression]];
NSExpression * left = [(NSComparisonPredicate *)parsed leftExpression];
calculatedResult = [left expressionValueWithObject:nil context:nil];
}
@catch (NSException *exception) {
NSLog(@"Input is not an expression...!");
}
@finally {
return calculatedResult;
}
}
it's simply start the expression with the operand "1.0 * " and everything will be floating point calculation onwards.
NSPredicate * parsed = [NSPredicate predicateWithFormat:[NSString stringWithFormat:@"1.0 * %@ = 0", expression]];
NB: Thanks @Martin R but, my question wasn't about integer division, it's totally about NSExpression. I have clearly excluded with my last sentence.
@Zaph, Exception handling is used here for a strong reasons. It is a place where my method is accepting user input wehre user can enter something like w*g and - expressionValueWithObject: context: will throw an exception and I have to avoid the abnormal termination of my App. If the user entered a valid expression then he/she will get an answer in the form of NSNumber otherwise a nil NSNumber object.
这篇关于如何从NSExpression的expressionValueWithObject：上下文方法获得浮点数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
