将java代码转换为Objective c

时间:2015-03-31 12:49:17

标签: java android ios objective-c iphone

我在.java文件中有几行代码, 我想要做的是我想将那行代码转换为客观的c行..

以下是行:

(1)
     return 9.87*Math.sin(Math.toRadians(2.0 * B)) - 7.53*Math.cos(Math.toRadians(B)) - 1.5*Math.sin(Math.toRadians(B));

(2)

    Calendar today = Calendar.getInstance();
    Calendar now = Calendar.getInstance();
    now.set(Calendar.DAY_OF_MONTH, 1);
    now.set(Calendar.MONTH, 0);
    now.set(Calendar.YEAR, today.get(Calendar.YEAR));
    (today.getTimeInMillis() - now.getTimeInMillis())/(1000*60*60*24.0) + 1;

有没有将此代码转换为客观c代码的在线工具? 还是请你帮我转换。

感谢您提前阅读并提前致谢

1 个答案:

答案 0 :(得分:1)

检查此代码或检查此参考链接of apple或查看此Stackoverflow link

NSDate *today = [NSDate date];
NSDate *now = [NSDate date];
NSCalendar *gregorian = [[NSCalendar alloc]
                         initWithCalendarIdentifier:NSCalendarIdentifierGregorian];
NSDateComponents *weekdayComponents =
[gregorian components:(NSDayCalendarUnit | NSWeekdayCalendarUnit) fromDate:today];
NSInteger day = [weekdayComponents day];
NSInteger weekday = [weekdayComponents weekday];

[today timeIntervalSince1970 ]-[now timeIntervalSince1970]/(1000*60*60*24.0)+1;

对于你的第一点使用这样的代码,但首先在你的.m文件#include <math.h>

中包含这个代码
return 9.87*sin((2.0 *B)*(M_PI/180.0)) - 7.53 * cos((B)*(M_PI/180.0)) - 1.5 * sin((B)*(M_PI/180.0));

我希望这段代码对你有用。