Question

我对objective-c很新。

我正在尝试创建一个需要英寸或厘米的函数，并将值转换为oposite公制系统。

鉴于我不知道如何相应地设置变量，我目前正在努力研究如何在objective-c中进行数学方程式。因为我总是从XCode得到一些错误，说变量是一个int但不应该是指针是错误的和其他东西......基本...我不知道如何制作数学......

目的是我传递存储在其他地方的字符串5’11”或1.80 cm并将其转换为相反的公制系统。

下面的代码尝试将英寸转换为cm ...如果我们传递5'5“，则数学等式应该是((5 * 12) + 3) * 2.54这等于180.34并且我看起来的格式化结果函数是1.80 cm

以下代码是我得到的，但它不适合我。如果有人能告诉我如何在我的代码中执行此操作，我将不胜感激。

- (NSString *)returnRightMetric:(NSString *)theMeasure typeOfMetricUsed:(NSString *)metricType
{
    if ([metricType isEqualToString:@"inches"]) {

    NSArray* theConvertion = [theMeasure componentsSeparatedByCharactersInSet:
                                [NSCharacterSet characterSetWithCharactersInString:@"’”"]];
    NSInteger* value1 = [theConvertion[0] intValue];
    NSInteger* value2 = [theConvertion[1] intValue];

    float *number = ((value1 * 12) + value2) * 2.54;
///.....

    } else if ([metricType isEqualToString:@"cm"])
    {
    }

    return result;
}

Answer 1

出现此问题的原因是您使用NSInteger和float中的指针。我已经使用cm和inches的实现修改了你的方法，如下所示：

- (NSString *)returnRightMetric:(NSString *)theMeasure typeOfMetricUsed:(NSString *)metricType
{
    NSString *result = nil;

    if ([metricType isEqualToString:@"inches"]) {
        NSArray* theConvertion = [theMeasure componentsSeparatedByCharactersInSet:
                                  [NSCharacterSet characterSetWithCharactersInString:@"’”"]];
        NSInteger value1 = [theConvertion[0] intValue];
        NSInteger value2 = [theConvertion[1] intValue];

        float number = ((value1 * 12) + value2) * 2.54;
        result = [NSString stringWithFormat:@"%.02f cm", round(number * 100.0) / 100.0];


    } else if ([metricType isEqualToString:@"cm"]) {
        float value = [theMeasure floatValue];
        float number = value / 2.54;

        if (number > 12.0) {
            result = [NSString stringWithFormat:@"%i’%i”", (int)floor(number / 12.0), (int)number % 12];

        } else {
            result = [NSString stringWithFormat:@"0’%i”", (int)round(number)];
        }

    }

    NSLog(@"result: %@", result);
    return result;
}

一些测试：

[self returnRightMetric:@"5’11”" typeOfMetricUsed:@"inches"];
[self returnRightMetric:@"180.34 cm" typeOfMetricUsed:@"cm"];
[self returnRightMetric:@"0’11”" typeOfMetricUsed:@"inches"];
[self returnRightMetric:@"27.94 cm" typeOfMetricUsed:@"cm"];

输出：

result: 180.34 cm
result: 5’11”
result: 27.94 cm
result: 0’11”

Answer 2

NSArray* theConvertion = [theMeasure componentsSeparatedByCharactersInSet:
                          [NSCharacterSet characterSetWithCharactersInString:@"’”"]];
int value1 = [theConvertion[0] intValue];
int value2 = [theConvertion[1] intValue];

float number = ((value1 * 12) + value2) * 2.54;
NSString *formattedNumber = [NSString stringWithFormat:@"%.02f", (number/100)];

简而言之，你需要知道原生/原始类型和引用类型之间的区别（使用指针这个字符的类型 - ＆gt; *。这可能有助于Link

Answer 3

从最后一个答案（来自tolgamorf）进行一些测试后，这里是最终版本。对我来说它完美无缺。

- (NSString *)returnRightMetric:(NSString *)theMeasure typeOfMetricUsed:(NSString *)metricType
{
    NSString *result = nil;

    if ([metricType isEqualToString:@"inches"]) {  
        if ([theMeasure isEqualToString:@""]) {
            return @"0";
        }
        NSArray* theConvertion = [theMeasure componentsSeparatedByCharactersInSet:
                                  [NSCharacterSet characterSetWithCharactersInString:@"’”"]];
        NSInteger value1 = [theConvertion[0] intValue];
        NSInteger value2 = [theConvertion[1] intValue];

        float number = ((value1 * 12) + value2) * 2.54;
        result = [NSString stringWithFormat:@"%.0f cm", round(number * 100.0) / 100.0];


    } else if ([metricType isEqualToString:@"cm"]) {
        float value = [theMeasure floatValue];
        float number = value / 2.54;

        if (roundf( number) >= 12.0) {
            if ((int)round( number) % 12==0) {
                result = [NSString stringWithFormat:@"%i’%i”", (int)roundf(number / 12.0), (int)round( number) % 12];
            }else{
                result = [NSString stringWithFormat:@"%i’%i”", (int)floorf(number / 12.0), (int)round( number) % 12];
            }
        } else {
            result = [NSString stringWithFormat:@"0’%i”", (int)round(number)];
        }
    }
    NSLog(@"result: %@", result);
    return result;
}

Answer 4

今天我们有来自HealthKit的有用API：

let cm_q = HKQuantity(unit: HKUnit(from: .centimeter), doubleValue: 180)
var inch_string = h_formatter.string(fromMeters: cm_q.doubleValue(for: HKUnit.meter()))

var inches_double = cm_q.doubleValue(for: HKUnit.inch())
let inch_q = HKQuantity(unit: HKUnit(from: .inch), doubleValue: inches_double)
var cm_d = inch_q.doubleValue(for: HKUnit(from: .centimeter))
var string = h_formatter.string(fromValue: cm_d, unit: .centimeter)

只需复制到Playground并播放值从cm转换为英寸并返回。格式化程序配置：

let h_formatter = LengthFormatter()
h_formatter.isForPersonHeightUse = true
h_formatter.unitStyle = .short

在结果中我们有：

180cm - ＆gt; “5'10.866”“并回到180cm

我希望它会有所帮助。

将英寸转换为厘米，反之亦然

4 个答案: