题
您将如何得到梦寐以求DRY理想在此示例中,在你选择的语言:
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:的自组织预处理宏众多可能是极其少可读可写所以这是出
解决方案
答案很简单:载体。 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);
其他提示
[我同意斯图尔特,但我按下作为学术锻炼。]
...刁钻
在某种程度上,你不是的真正的重复自己;你正在做的两个类似的东西,是(名副其实地)相互正交的。
我想你可以做到以下几点,虽然它没有更多的可读性,当然也没有更好的性能:
[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);
}
在第二个想法,我认为这仅仅是一个愚蠢的想法...: - )
也许,在这种情况下,你正在干太多了极致?
然而,作为一项学术活动,我期待着看到是否有人能拿出一个解决方案,作为当前的代码的可读性 - 但没有明显重复
!如果您的网格是方形的,我觉得有以下可以工作:
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));
}
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