Comments (3)
我想这里不需要改,因为这里判断的是首个或者第二个字符即是0的情况,也就是“-0”,或者“0”,那么就直接返回。这里假设十进制整数的第一个数字不能为0(因为这通常用于表达8进制)
from mql4-lib.
但若數據為負數,最後會變成正數,例如 -0.6 會變成 0.6。
下面這邊先標記為負數了,並取下一個字元
if(c=='-')
{
sign=-1;
c=m_stream.nextChar();
}
以-0.6為例的話,會進入上面if內,標記為負數,並取下一個字元0
然後繼續走到
if(c=='0') // 0
{
sign=1; // 這裡做sign=1會把負數變為正
return true;
}
進入if後 sign=1 又把它改回正數了,
然後退出parseInt這個方法,
走到parseNumber最下面的
value=sign*(intValue+fracValue)*MathPow(10,expValue);
-0.6 就變成 0.6了
from mql4-lib.
我看了一下代码,这个 parseInt 并不是独立的,是用在 parseNumber 中的,因此确实是有问题的。多谢~
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