我正在試驗在Xcode IDE中用Python編寫的Hilbert曲線。代碼清單是:python打印語法錯誤
# python code to run the hilbert curve pattern
# from http://www.fundza.com/algorithmic/space_filling/hilbert/basics/index.html
import sys, math
def hilbert(x0, y0, xi, xj, yi, yj, n):
if n <= 0:
X = x0 + (xi + yi)/2
Y = y0 + (xj + yj)/2
# Output the coordinates of the cv
print '%s %s 0' % (X, Y)
else:
hilbert(x0, y0, yi/2, yj/2, xi/2, xj/2, n - 1)
hilbert(x0 + xi/2, y0 + xj/2, xi/2, xj/2, yi/2, yj/2, n - 1)
hilbert(x0 + xi/2 + yi/2, y0 + xj/2 + yj/2, xi/2, xj/2, yi/2, yj/2, n - 1)
hilbert(x0 + xi/2 + yi, y0 + xj/2 + yj, -yi/2,-yj/2,-xi/2,-xj/2, n - 1)
def main():
args = sys.stdin.readline()
# Remain the loop until the renderer releases the helper...
while args:
arg = args.split()
# Get the inputs
pixels = float(arg[0])
ctype = arg[1]
reps = int(arg[2])
width = float(arg[3])
# Calculate the number of curve cv's
cvs = int(math.pow(4, reps))
# Begin the RenderMan curve statement
print 'Basis \"b-spline\" 1 \"b-spline\" 1'
print 'Curves \"%s\" [%s] \"nonperiodic\" \"P\" [' % (ctype, cvs)
# Create the curve
hilbert(0.0, 0.0, 1.0, 0.0, 0.0, 1.0, reps)
# End the curve statement
print '] \"constantwidth\" [%s]' % width
# Tell the renderer we have finished
sys.stdout.write('\377')
sys.stdout.flush()
# read the next set of inputs
args = sys.stdin.readline()
if __name__ == "__main__":
main()
我得到在Xcode以下錯誤: 文件 「/Users/248239j/Desktop/hilbert/hilbertexe.py」,第12行 打印 '%s%S' %( X,Y) ^ 語法錯誤:無效的語法
是否有人有該代碼行的替代方案?提前致謝。我今天開始使用python。
可能的複製(http://stackoverflow.com/questions/826948/語法錯誤的打印與蟒蛇-3) – tripleee