画线WINAPI

时间:2017-04-08 16:10:13

标签: c++ winapi

我想通过点击并从一点到另一点移动光标来绘制一条线,我刚刚复制了以下代码,而我的WindowProcedure看起来就是这样:

LRESULT CALLBACK WindowProc(HWND hwnd, UINT uMsg, WPARAM wParam, LPARAM lParam)
{   
    //PAINTSTRUCT ps;
    HDC hdc;
    bool draw = false;
    int x = 0, y=0;

    //InvalidateRect(hwnd, NULL, true);
    switch (uMsg)
    {
    case WM_LBUTTONDOWN:
            draw = true;
            x = LOWORD(lParam);
            y = HIWORD(lParam);
            return 0;

    case WM_LBUTTONUP:
        if (draw)
        {
            hdc = GetDC(hwnd);
            MoveToEx(hdc, x, y, NULL);
            LineTo(hdc, LOWORD(lParam), HIWORD(lParam));
            ReleaseDC(hwnd, hdc);
        }
        draw = FALSE;
        return 0;

    case WM_MOUSEMOVE:
        if (draw)
        {
            hdc = GetDC(hwnd);
            MoveToEx(hdc, x, y, NULL);
            LineTo(hdc, x = LOWORD(lParam), y = HIWORD(lParam));
            ReleaseDC(hwnd, hdc);
        }
        return 0;


    case WM_DESTROY:
        PostQuitMessage(0);
        return 0;

    }
    return DefWindowProc(hwnd, uMsg, wParam, lParam);
   }

但是当我点击没有任何反应时,就像只有一个案件可以处理,是否正确?当我评论前两种情况时它会绘制线条,所以它进入切换但不是我想要做的。有什么建议吗?

1 个答案:

答案 0 :(得分:-2)

LRESULT CALLBACK WindowProc(HWND hwnd, UINT uMsg, WPARAM wParam, LPARAM lParam)
{   
    //PAINTSTRUCT ps;
    HDC hdc;
    static bool draw = false;  // ←static
    static int x = 0, y=0;     // ←

这个怎么样?