中西文化
差异
导致在电子信息里处理也大不相同在英文里只需要26个字母就可以显示所有文章了而在中文里需要最基本
就有2000多个
对于
些在嵌入式软件Software里要显示那么就得手动去构造所有图形这是个比较大工作量如果让每个厂家都去完成这个任务显然是不可能面对着大量嵌入式用户需求那么就需要解决中文字模图形问题毕竟大家经常使用Windows最先想到肯定是如何样把里面提取图形出来生成自己需要几个字库下面就来介绍如何样用
GetGlyphOutline获取显示图形数据GetGlyphOutline声明如下:WINGDIAPI DWORD WINAPI GetGlyphOutlineA( __in HDC hdc,
__in UINT uChar,
__in UINT fuFormat,
__out LPGLYPHMETRICS lpgm,
__in DWORD cjBuffer,
__out_bcount_opt(cjBuffer) LPVOID pvBuffer,
__in CONST MAT2 *lpmat2
);
WINGDIAPI DWORD WINAPI GetGlyphOutlineW( __in HDC hdc,
__in UINT uChar,
__in UINT fuFormat,
__out LPGLYPHMETRICS lpgm,
__in DWORD cjBuffer,
__out_bcount_opt(cjBuffer) LPVOID pvBuffer,
__in CONST MAT2 *lpmat2
);
#
def UNICODE
#
GetGlyphOutline GetGlyphOutlineW
#
# GetGlyphOutline GetGlyphOutlineA
#end // !UNICODEhdc是设备句柄
uChar是需要获取图形数据
fuFormat是获取数据
格式lpgm是获取
相关信息cjBuffer是保存
数据缓冲区大小pvBuffer是保存
数据缓冲区lpmat2是3*3
变换矩阵
例子如下:#001 //浮点数据转换为固定浮点数
#002 FIXED FixedFromDouble(double d)
#003 {
#004 long l;
#005 l = (long) (d * 65536L);
#006 
*(FIXED *)&l;
#007 }
#008
#009 //设置字体图形变换矩阵
#010 void SetMat(LPMAT2 lpMat)
#011 {
#012 lpMat->eM11 = FixedFromDouble(2);
#013 lpMat->eM12 = FixedFromDouble(0);
#014 lpMat->eM21 = FixedFromDouble(0);
#015 lpMat->eM22 = FixedFromDouble(2);
#016 }
#017
#018 //
#019 //获取字模信息
#020 //蔡军生 2007/12/16 qq:9073204 深圳
#021 void TestFontGlyph(void)
#022 {
#023 //创建字体
#024 HFONT hFont = GetFont
;
#025
#026 //设置字体到当前设备
#027 HDC hDC = ::GetDC(m_hWnd);
#028 HFONT hOldFont = (HFONT)SelectObject(hDC,hFont);
#029
#030 //设置字体图形变换矩阵
#031 MAT2 mat2;
#032 SetMat(&mat2);
#033
#034
#035 GLYPHMETRICS gm;
#036
#037 //设置要显示
#038 TCHAR chText = L'蔡';
#039
#040 //获取这个图形需要字节大小
#041 DWORD dwNeedSize = GetGlyphOutline(hDC,chText,GGO_BITMAP,&gm,0,NULL,&mat2);
#042 (dwNeedSize > 0 && dwNeedSize < 0xFFFF)
#043 {
#044 //按需要分配内存
#045 LPBYTE lpBuf = (LPBYTE)HeapAlloc(GetProcessHeap,HEAP_ZERO_MEMORY,dwNeedSize);
#046 (lpBuf)
#047 {
#048 //获取图形数据到缓冲区
#049 GetGlyphOutline(hDC,chText,GGO_BITMAP,&gm,dwNeedSize,lpBuf,&mat2);
#050
#051 //计算图形每行占用字节数
#052 
nByteCount = ((gm.gmBlackBoxX +31) >> 5) << 2;
#053
#054 //显示每行图形数据
#055 for ( i = 0; i < gm.gmBlackBoxY; i
)
#056 {
#057 //
#058 for ( j = 0; j < nByteCount; j)
#059 {
#060
#061 BYTE btCode = lpBuf[i* nByteCount + j];
#062
#063 //按字节输出每点数据
#064 for ( k = 0; k < 8; k)
#065 {
#066
#067 (btCode & (0x80>>k))
#068 {
#069
#070 OutputDebugString(_T("1"));
#071 }
#072
#073 {
#074 OutputDebugString(_T("0"));
#075 }
#076
#077 }
#078
#079 }
#080
#081 //
#082 OutputDebugString(_T("\r\n"));
#083 }
#084
#085 //
#086 HeapFree(GetProcessHeap,0,lpBuf);
#087 }
#088 }
#089
#090 //
#091 SelectObject(hDC,hOldFont);
#092 DeleteObject(hFont);
#093
#094 //
#095 ReleaseDC(m_hWnd,hDC);
#096 }
#097 输出
结果如下:00000000000000010000000000000000
00000000110000011000000000000000
00000000100000011000000000000000
00000000100000011000011000000000
11111111111111111111111100000000
01000000100000011000000000000000
00000000100000011000000000000000
00000100100000011000000000000000
00000110100000010000000000000000
00000100000000000000000000000000
00001100000001000000100000000000
00001111111101111111110000000000
00001000001111000000110000000000
00011000001000100001100000000000
00010100001000100001000000000000
00010010011000100011000000000000
00100011010000010010000000000000
00110010110000011010000000000000
01011000110000001100000000000000
10001000100000001100000000000000
00001001100000010110000000000000
00001011011111111011000000000000
00000010000000000001110000000000
00000110000000000000111100000000
00001100000000000110011100000000
00011111111111111111001000000000
00110000000010000000000000000000
01000000000010000000000000000000
00000001000010000000000000000000
00000011100010001100000000000000
00000011000010000010000000000000
00000110000010000011100000000000
00001100000010000001100000000000
00011000100010000000110000000000
00110000011110000000110000000000
01000000001110000000010000000000
00000000000100000000000000000000
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