您将如何以您选择的语言在此示例中获得梦想的DRY理想:
drawLine(Point(0, 0), Point(w, 0));
int curRowY = 0;
for(int row=0; row<rowHeights.size(); row++) {
curRowY += rowHeights[row];
drawLine(Point(0, curRowY), Point(w, curRowY));
}
drawLine(Point(0, 0), Point(0, h));
int curColX = 0;
for(int col=0; col<colWidths.size(); col++) {
curColX += colWidths[col];
drawLine(Point(curColX, 0), Point(curColX, h));
}
注意:大量的临时预处理器宏可能非常 可读和可写,因此无法使用。
答案 0 :(得分:9)
答案很简单:矢量。 E.g。
repeatLines(Point start, Point end, Vector direction, int[] gaps)
{
drawLine(start, end);
for (int i = 0; i < gaps.Length; i++)
{
Vector vector = direction * gaps[i];
start += vector;
end += vector;
drawLine(start, end);
}
}
repeatLines(Point(0, 0), Point(0, w), Vector(1, 0), rowHeights);
repeatLines(Point(0, 0), Point(h, 0), Vector(0, 1), colWidths);
答案 1 :(得分:1)
[我同意斯图尔特的观点,但我继续作为学术演习。]
...刁钻
在某种程度上,你不是真的重复自己;你正在做两个相似的事情(字面上和比喻上)彼此正交。
我想你可以做到以下几点,虽然它不再具有可读性,当然也没有更高的性能:
[伪C#]:
void DrawGrid()
{
DrawLines(w, rowHeights, true);
DrawLines(h, colWidths, false);
}
void DrawLines(int lineLength, int[] lineSeparations, bool isHorizontal)
{
MyDrawLine(Point(0, 0), Point(lineLength, 0), isHorizontal);
int offset = 0;
for (int i = 0; i < widths.length; i++)
{
offset += lineSeparations[i];
MyDrawLine(Point(offset, 0), Point(offset, lineLength), isHorizontal);
}
}
void MyDrawLine(Point startPoint, Point endPoint, bool isHorizontal)
{
if (isHorizontal)
{
SwapXAndYCoordinates(startPoint);
SwapXAndYCoordinates(endPoint);
}
drawLine(startPoint, endPoint);
}
再想一想,我认为这只是一个愚蠢的想法......: - )
答案 2 :(得分:0)
也许,在这种情况下,你将DRY带到了极端的极端?
尽管如此,作为一项学术活动,我期待看到是否有人能够提出与您当前代码一样可读的解决方案 - 但没有明显的重复!
答案 3 :(得分:0)
如果您的网格是方形的,我认为以下方法可行:
void drawGrid()
{
for(int i = 1, offset = 10; i <= numPoints; i++, offset += 10)
{
Point p = new Point(i * offset, i * offset);
drawHorizontal(p);
drawVertical(p);
}
}
void drawHorizontal(Point p)
{
drawLine(new Point(0, p.y), new Point(width, p.y));
}
void drawVertical(Point p)
{
drawLine(new Point(p.x, 0), new Point(p.x, height));
}