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编译不过
增加: template inline bool operator()(const TIndex& x, const TIndex& y)const { TDigest di(y); return (x + i<n) ? (blocks[x + i]<di.value) : true; } 这种函数。。目前编译过了
Originally posted by @codeman001 in #36 (comment)
The text was updated successfully, but these errors were encountered:
谢谢你的问题, 你的实现有误;
我试过vc、xcode、gcc、clang编译过; 我猜测是: 表达式原地构造的对象不能当作 const T& 参数被传递;但看错误说明又不像:( 你可以试试 把它先定义成一个变量 from #36 (comment)
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我用的vs2010测试; 看来vs2013库的实现有些不同,已经调整代码; from #36 (comment)
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编译不过
增加:
template
inline bool operator()(const TIndex& x, const TIndex& y)const {
TDigest di(y);
return (x + i<n) ? (blocks[x + i]<di.value) : true;
}
这种函数。。目前编译过了
Originally posted by @codeman001 in #36 (comment)
The text was updated successfully, but these errors were encountered: