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我有一个包含多个数字和运算符的字符串

@ "5+4-9/10".

如何从中得到结果?

我想在我正在使用的计算器中使用它。当按下数字或运算符时,我将不得不动态显示结果。

我在android中使用了 arity jar 文件。但我无法在iPhone中实现类似的目标。

4

3 回答 3

9

那这个呢 -

NSString *formula = @"1+5*6";
NSExpression *exp = [NSExpression expressionWithFormat:formula];
NSNumber *resultForCustomFormula = [exp expressionValueWithObject:nil context:nil];
NSLog(@"%f", [resultForCustomFormula floatValue]);

编辑: 现在我考虑了您的要求并使用了NSScanner您不会相信在Borrrden 先生NSScanner建议我使用它之前我没有使用它并且我发现它很棒。见下面的方法 -

-(NSMutableString *)formatString:(NSString *)formula
{
    // Let's check if there any wrong (.) value exm: 1/.2 or .7+3 
    // 1/0.2 and 0.7+3 are okay but above are incorrect so first fix them

    NSString *str = formula;
    NSInteger c = 0;
    for(int i=0; i<[str length]; i++)
    {

        if([[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"+"] ||
           [[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"-"] ||
           [[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"/"] ||
           [[NSString stringWithFormat:@"%c",[str characterAtIndex:i]] isEqualToString:@"*"])
        {
            if([str length] > i+1)
            {
                if([[NSString stringWithFormat:@"%c",[str characterAtIndex:i+1]] isEqualToString:@"."])
                {
                    formula = [formula stringByReplacingCharactersInRange:NSMakeRange(i+1+c, 1) withString:@"0."];
                    c++;
                }
            }
        }
    }

    // Now we will convert all numbers in float

    NSString *aString;
    float aFloat;
    NSMutableString *formattedString = [[NSMutableString alloc]init];

    NSScanner *theScanner = [NSScanner scannerWithString:formula];
    while ([theScanner isAtEnd] == NO) 
    {

        if([theScanner scanFloat:&aFloat])
        {
            [formattedString appendString:[NSString stringWithFormat:@"%f",aFloat]];
        }

        if([theScanner scanUpToCharactersFromSet:[NSCharacterSet decimalDigitCharacterSet] intoString:&aString])
        {
            [formattedString appendString:aString];
        }
    }
    return formattedString;
}

这会将(2.222/.4)+9999-7+0.7*.13转换为(2.222000/0.400000)+9999.000000-7.000000+0.700000*0.130000
只需在使用之前调用此方法NSExpression

NSString *formula = @"(2.222/.4)+9999-7+0.7*.13";
NSString *formattedString = [self formatString:formula];
NSExpression *exp = [NSExpression expressionWithFormat:formattedString];
NSNumber *resultForCustomFormula = [exp expressionValueWithObject:nil context:nil];
NSLog(@"Result = %f", [resultForCustomFormula floatValue]);

//OutPut: Result = 9997.646484

注意:我并不是说它适用于所有公式字符串。可能在某些情况下它不起作用。但它适用于一般方程。

于 2013-01-04T10:50:34.430 回答
3

Note that : 您必须至少传递一个浮点值

我知道这不是最好的方法,但是这样的东西可以为你工作。

NSString *formula = @"5+4-9/10";
NSString *str = [formula lastPathComponent];
formula = [formula stringByReplacingOccurrencesOfString:str withString:[NSString stringWithFormat:@"%@.0",str]];
NSString *strCal=[NSString stringWithString:formula];
NSExpression *exp=[NSExpression expressionWithFormat:strCal];
float result=[[exp expressionValueWithObject:nil context:nil] floatValue];
NSLog(@"result:%f",result);

您可以使用GCMathParserDDMathParser

不知道这是否是最有效的方法,但还是写来帮助你......!!!

于 2013-01-04T10:37:00.183 回答
0

您可以将“所有”数字解析为浮动:

NSMutableString *numericExpression = [NSMutableString stringWithString:@"4+3/2-7/4"];

NSMutableString *nExp=[NSMutableString stringWithString:@""];
BOOL innum=false;
BOOL dotfound=false;

for (NSInteger charIdx=0; charIdx<numericExpression.length; charIdx++){
        NSString *substr=[NSString stringWithFormat:@"%C",[numericExpression characterAtIndex:charIdx]];
        if (isdigit([numericExpression characterAtIndex:charIdx])) {
            // in number
            innum=true;
        }else if ([substr isEqualToString:@"."]){
            // dot found ...
            innum=true;
            dotfound=true;
        }else{
            // not in number
            if (innum && dotfound) {
                NSLog(@"all good (*).*");
            }else if(innum && !dotfound){
                NSLog(@"adding .0");
                [nExp appendString:@".0"];
            }
            innum=false;
            dotfound=false;
        }
    [nExp appendString:substr];
    NSLog(@"%@",nExp);
}

NSExpression *expression = [NSExpression expressionWithFormat:nExp];
NSNumber *result = [expression expressionValueWithObject:nil context:nil];
 NSLog(@"calc %@ = %@",nExp,result);
于 2016-11-15T16:12:25.650 回答