vec*_*r07 1 c++ unordered-map segmentation-fault
当datact = 10736尝试插入unordered_map时,会发生分段错误(请参阅调用行,该调用行将调用该错误).请参阅下面的尝试修复.
当抛出SIGSEGV时,它指向我的第764行 hashtable_policy.h
INPUT:数据文件,其中column1 = count,column2 = 16个字符的字符串
目的:通过将1-替换不同序列的所有计数加在一起来聚类16个字符的序列.看到的第一个序列是"起源",通过它来识别其所有的1个替换朋友.
PSEUDOCODE:对于文件中的每一行:
读取计数,读取序列.
如果序列key_value存在于散列'location'(类型unordered_map)中,则添加当前计数;
否则创建一个新的key_value,让它指向此处的计数,并指定所有1个替换序列也指向此计数.
码:
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
#include <cmath>
#include <vector>
#include <unordered_map>
#include <map>
#include "matrix.h"
using namespace std;
int nuc2num(char currchar)
{ // returns 0,1,2,3 for A,C,T,G respectively
int outnum;
if (currchar=='A')
{
outnum=0;
}
else if (currchar=='C')
{
outnum=1;
}
else if (currchar=='T')
{
outnum=2;
}
else
{
outnum=3;
}
return outnum;
}
int main(int argc, char* argv[])
{
//command line arguments
// arg1=filename, arg2=barcode sequence length, arg3=#mismatches permitted
//input handling
// file format: column 1 | column 2
// counts | sequence [int | string]
string filename;
string funnelstring;
// define lookup matrix; rows=ACTG, cols = ACTG without row element
Matrix <char> sub_lookup(4,3);
sub_lookup[0][0] = 'C';
sub_lookup[0][1] = 'T';
sub_lookup[0][2] = 'G';
sub_lookup[1][0] = 'A';
sub_lookup[1][1] = 'T';
sub_lookup[1][2] = 'G';
sub_lookup[2][0] = 'A';
sub_lookup[2][1] = 'C';
sub_lookup[2][2] = 'G';
sub_lookup[3][0] = 'A';
sub_lookup[3][1] = 'C';
sub_lookup[3][2] = 'T';
int L,k;
int j=0;
const int buffersize=10000;
int currentsize=buffersize;
int datact=0;
int currchar;
vector <unsigned int> ctarr(buffersize);
vector <string> seqarr(buffersize);
filename=argv[1];
L=atoi(argv[2]);
k=atoi(argv[3]);
unsigned int sct;
int substrlen;
string sequence,textct;
ifstream seqfile (filename.c_str());
//map <string,unsigned int*> location;
unordered_map <string,unsigned int*> location;
if (seqfile.is_open())
{
getline(seqfile,textct,'\n');
while (textct != "")
{
sct=atoi(textct.c_str());
substrlen=textct.length();
//cout << textct << endl;
sequence=textct.substr(substrlen-L,L);
//cout << sequence << endl;
//is there an associated guy?
if (location.find(sequence) != location.end()) //just asks whether this key has been assigned
{ //there's a value in the region
*location[sequence]+=sct;
}
else
{ //no value in region, make a footprint
ctarr[datact]=sct;
seqarr[datact]=sequence;
location[sequence]=&ctarr[datact]; //assign current key to point to data count
//assign k substitution "funnel" region to point to this count as well
for (j=0; j<L; j++)
{
funnelstring=sequence;
currchar = nuc2num(sequence[j]);
if (datact==10736 && j==13)
{
cout << "here" << endl;
cout << sequence << endl;
}
for (k=0; k<3; k++)
{
funnelstring[j]=sub_lookup[currchar][k];
// if (datact==10736 && j==13)
// {
// cout << funnelstring << endl;
// cout << location.max_size() << " | " << location.size() << endl;
// string asdf;
// asdf="AAAAAAAAAAAAAAAA";
// location[asdf]=&ctarr[datact]; //still segfaults
// }
if (location.find(funnelstring) == location.end()) // asks whether this key has been assigned
{ //this region is not assigned to another funnel
location[funnelstring]=&ctarr[datact]; //LINE THAT CAUSES SIGSEGV
}
}
}
datact++;
cout << datact << endl;
if (datact>=currentsize)
{
ctarr.resize(currentsize+buffersize);
seqarr.resize(currentsize+buffersize);
currentsize+=buffersize;
}
}
getline(seqfile,textct,'\n');
}
seqfile.close();
}
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探索.
datact==10736和j=13,任意键添加到一个SIGSEGV(unordered_map)定位结果.datact= = 16-35千).如果需要,我可以发布读取的数据文件.
编辑:解决
而不是unordered_map <string,unsigned int*> location,替换为unordered_map <string,unsigned int> location(value_type是int而不是int*).现在value_type保存ctarr []中的索引.运行很好.谢谢!
vector调用时,元素的指针可以无效vector::resize().这是因为可能必须移动整个数据以便找到适合新大小的连续内存块.换句话说:一旦你打电话resize,你的所有location数据都会突然变成无用的垃圾.
可能的解决方案:
location将所需元素的索引存储ctarr为其值而不是指针.(这肯定不会改变你程序的语义.)location存储实际unsigned int 值而不是指针.根据您的程序逻辑以及更改和访问此数据的方式,这可能不是您想要的.另请注意,虽然发生了段hashtable_policy.h错误,但这个错误与unordered_map(或vector)的实现无关- 完全没有阅读参考文献的错误;-) vector::resize():http://www.cplusplus.com/reference/ vector/vector/resize /("Iterator validity"部分)
我注意到你的代码的另一件事是你operator[]用来访问你的vector元素.这会禁用越界检查.如果我在我的代码遇到像你的错误来了(很难,因为它从我的错误码远的地方发生追溯),我的首要行动将是交换operator[]的vector::at()(实际上,我总是先从at()只有开关,如果我能证明毫无疑问,边界检查是这个特定目的的性能瓶颈).这对您的问题没有帮助,但通常是发现错误的宝贵帮助.
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