当前位置: 首页>>代码示例>>C++>>正文


C++ path_t::clear方法代码示例

本文整理汇总了C++中path_t::clear方法的典型用法代码示例。如果您正苦于以下问题:C++ path_t::clear方法的具体用法?C++ path_t::clear怎么用?C++ path_t::clear使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在path_t的用法示例。


在下文中一共展示了path_t::clear方法的2个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C++代码示例。

示例1: as

void
export_md_to_apath(const char* md,
                   const bool is_fwd_strand,
                   path_t& apath,
                   const bool is_edge_deletion_error)
{

    // to make best use of previous code, we parse the MD in the
    // alignment direction and then orient apath to the forward strand
    // as a second step if required
    //
    assert(NULL != md);

    apath.clear();
    export_md_to_apath_impl(md,apath);

    unsigned as(apath.size());

    if ( ((as>0) and (apath.front().type == DELETE)) or
         ((as>1) and (apath.back().type == DELETE)) )
    {
        std::ostringstream oss;
        if (is_edge_deletion_error)
        {
            oss << "ERROR: ";
        }
        else
        {
            oss << "WARNING: ";
        }
        oss << "alignment path: " << apath_to_cigar(apath) << " contains meaningless edge deletion.\n";
        if (is_edge_deletion_error)
        {
            throw blt_exception(oss.str().c_str());
        }
        else
        {
            log_os << oss.str();
            path_t apath2;
            for (unsigned i(0); i<as; ++i)
            {
                if (((i==0) or ((i+1)==as)) and
                    apath[i].type == DELETE) continue;
                apath2.push_back(apath[i]);
            }
            apath=apath2;
            as=apath.size();
        }
    }

    if ( (not is_fwd_strand) and (as>1) )
    {
        std::reverse(apath.begin(),apath.end());
    }
}
开发者ID:ctb,项目名称:quast,代码行数:55,代码来源:align_path_match_descriptor.cpp

示例2: cptr

void
cigar_to_apath(const char* cigar,
               path_t& apath)
{
    using illumina::blt_util::parse_unsigned;

    assert(NULL != cigar);

    apath.clear();

    path_segment lps;
    const char* cptr(cigar);
    while (*cptr)
    {
        path_segment ps;
        // expect sequences of digits and cigar codes:
        if (! isdigit(*cptr)) unknown_cigar_error(cigar,cptr);
        ps.length = parse_unsigned(cptr);
        ps.type = cigar_code_to_segment_type(*cptr);
        if (ps.type == NONE) unknown_cigar_error(cigar,cptr);
        cptr++;
        if ((ps.type == PAD) || (ps.length == 0)) continue;

        if (ps.type != lps.type)
        {
            if (lps.type != NONE) apath.push_back(lps);
            lps = ps;
        }
        else
        {
            lps.length += ps.length;
        }
    }

    if (lps.type != NONE) apath.push_back(lps);
}
开发者ID:ctb,项目名称:quast,代码行数:36,代码来源:align_path.cpp


注:本文中的path_t::clear方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。