87 lines
		
	
	
		
			2.3 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
			
		
		
	
	
			87 lines
		
	
	
		
			2.3 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
| // For more tutorials: https://blog.learngoprogramming.com
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| //
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| // Copyright © 2018 Inanc Gumus
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| // Learn Go Programming Course
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| // License: https://creativecommons.org/licenses/by-nc-sa/4.0/
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| //
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| 
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| package main
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| 
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| import (
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| 	"fmt"
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| 	"time"
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| 
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| 	"github.com/inancgumus/screen"
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| )
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| 
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| func main() {
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| 	for shift := 0; ; shift++ {
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| 		// we need to clear the screen here.
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| 		// or the previous character will be left on the screen
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| 		//
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| 		// alternative: you can fill the rest of the missing placeholders
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| 		//              with empty lines
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| 		screen.Clear()
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| 		screen.MoveTopLeft()
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| 
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| 		now := time.Now()
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| 		hour, min, sec := now.Hour(), now.Minute(), now.Second()
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| 
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| 		clock := [...]placeholder{
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| 			digits[hour/10], digits[hour%10],
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| 			colon,
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| 			digits[min/10], digits[min%10],
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| 			colon,
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| 			digits[sec/10], digits[sec%10],
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| 		}
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| 
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| 		for line := range clock[0] {
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| 			l := len(clock)
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| 
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| 			// this sets the beginning and the ending placeholder positions (indexes).
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| 			// shift%l prevents the indexing error.
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| 			s, e := shift%l, l
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| 
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| 			// to slide the placeholders from the right part of the screen.
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| 			//
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| 			// here, we assume that as if the clock's length is double of its length.
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| 			// this makes things easy to manage: that's why: l*2 is there.
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| 			//
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| 			// shift is always increasing, for it's to go beyond the clock's length,
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| 			// it should be equal or greater than l*2, right (after the remainder of course)?
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| 			//
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| 			// so, if the clock goes beyond its length; this code detects that,
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| 			// and resets the starting position to the first placeholder's index,
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| 			// and it keeps doing so until the clock is fully displayed again.
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| 			if shift%(l*2) >= l {
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| 				s, e = 0, s
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| 			}
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| 
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| 			// print empty lines for the missing place holders.
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| 			// this creates the effect of moving placeholders from right to left.
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| 			//
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| 			// l-e can only be non-zero when the above if statement runs.
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| 			// otherwise, l-e is always zero, because l == e.
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| 			//
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| 			// this is one of the other benefits of assuming the length of the
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| 			// clock as the double of its length. otherwise, l-e would always be 0.
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| 			for j := 0; j < l-e; j++ {
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| 				fmt.Print("     ")
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| 			}
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| 
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| 			// draw the digits starting from 's' to 'e'
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| 			for i := s; i < e; i++ {
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| 				next := clock[i][line]
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| 				if clock[i] == colon && sec%2 == 0 {
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| 					next = "   "
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| 				}
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| 
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| 				fmt.Print(next, "  ")
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| 			}
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| 			fmt.Println()
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| 		}
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| 
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| 		time.Sleep(time.Second)
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| 	}
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| }
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