这是我的问题。我在游戏中为我的Gui制作了一个TextBox。
它的作用是,每当我调整它的大小时,因为它包装了文字,我必须弄清楚文本字符串中插入符号所在的索引,然后我需要在reize之后将其转换为正确的行列。根据我的分析器最慢的部分是当我得到下一个要评估的unicode字符时:
int AguiTextBox::indexFromColumnRow( int column, int row, bool includeUnwantedChars ) const
{
size_t rowLen = 0;
int retIndex = -1;
int bytesSkipped = 0;
int curCharLen = 0;
std::string curChar;
std::string::const_iterator it = getText().begin();
std::string::const_iterator end = getText().end();
//decrement column so that the lowest is -1
column--;
if(textRows.size() == 0 || (column == -1 && row == 0))
{
//not in the text
return -1;
}
0.01s for(size_t i = 0; i < textRows.size(); ++i)
{
//get length of row
0.00s rowLen = _unicodeFunctions.getUtf8StringLength(textRows[i]);
//handle -1th case
//get next character
do
{
0.00s curCharLen = _unicodeFunctions.bringToNextUnichar(it,end);
0.01s curChar = getText().substr(bytesSkipped,curCharLen);
bytesSkipped += curCharLen;
if(includeUnwantedChars)
retIndex++;
} while (curChar[0] >= 0 && curChar[0] < ' ' && curChar != "\n");
if(!includeUnwantedChars)
retIndex++;
//only increase for newlines
0.00s if(curChar != "\n")
{
bytesSkipped -= curCharLen;
retIndex--;
it -= curCharLen;
}
if((int)i == row && column == -1)
{
return retIndex;
}
0.06s for(size_t j = 0; j < rowLen; ++j)
{
//get next character
do
{
0.10s curCharLen = _unicodeFunctions.bringToNextUnichar(it,end);
0.91s curChar = getText().substr(bytesSkipped,curCharLen);
0.03s bytesSkipped += curCharLen;
0.03s if(includeUnwantedChars)
retIndex++;
0.11s } while (curChar[0] >= 0 && curChar[0] < ' ' && curChar != "\n");
0.06s if(!includeUnwantedChars)
0.00s retIndex++;
0.02s if((int)i == row && (int)j == column)
{
return retIndex;
}
}
}
return retIndex;
}
我该如何优化呢?
由于
@Erik对字符的双端队列意味着什么?
答案 0 :(得分:1)
您正在使用以下内容提取子字符串:
curChar = getText().substr(bytesSkipped,curCharLen);
但是你只使用第一个元素。您只需提取所需的char
即可避免字符串构造/复制。
在一般算法优化方面 - 我将花费构建deque
个角色对象所需的资源,而不是使用std::string
。这将允许您直接索引任何字符,无需一遍又一遍地扫描和解析相同的utf-8序列。