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我想从之前的函数值中画出一条线。因此,我需要一个返回函数,所以在图形函数中绘制一条线更为顺利。为什么仍然有错误返回类型需要?
我很困惑,为什么仍然有错误,因为我把它放回去的函数。请帮我..
:;
public static SLL(int SLL)
public Program()
{
super("MyProgram"); //window name
.....
computeButton.addActionListener(this);
}
public void actionPerformed(ActionEvent ee)
{
}
public static SLL(int SLL)
{ ....
HomePosition HP[][] = new HomePosition[2][6];
HomePosition DP[][]= new HomePosition[2][6];
HomePosition FB[][] =new HomePosition[2][6];
...
FB[1][0].x= FB[2][0].x=40;
FB[1][0].y=FB[2][0].y=40;
HP[1][0].x = 40; HP[1][0].y = 30;
HP[1][1].x = 26; HP[1][1].y = 30;
DP[1][0].x = 30; DP[1][0].y =40;
DP[1][1].x = 16; DP[1][1].y = 40;
SLL.SLx[m][n]=HP[m][n].x;
return SLL;
}
public void paint(Graphics g) {
super.paint(g);
}
BasicStroke stroke;
Graphics2D g2 = (Graphics2D)g;
stroke = new BasicStroke(4.0f);
g2.setStroke(stroke);
.....
for (m=1;m<=2;m++){
for (n=0;n<=1;n++){
g2.drawLine(SLL.SLx[m][n]*M,40*M,40*M,30*M);
}
}
}
public static void main(String[] arguments)
{
Program frame=new Program();
final int WIDTH=910;
final int LENGTH=725;
frame.setSize(WIDTH,LENGTH);
final int buttonWidth=30;
final int buttonHeight=10;
frame.setVisible(true);
}
}
考虑使用更多的描述性变量名称。最后我检查了几十年前他们停止收费。 –
如果你有'SLL.SLx [m] [n] ...'' –
@MattBall是正确的,SLL是一个'int'。特别是当你有一个方法名和一个相同的变量名。哇! –