CODEWITHSUNDEEP
c++vectornew-operator
Tex
Tex

Reputation: 950

STL Vectors and the new operator

This question should be quite straightforward, maybe stupid but I just can't find the problem.

Basically, I have to parse some sentences in natural language. I need to implement a simple algorithm that manipulates "Blocks". A Block is made of 2 Pseudosentences, which are made of 20 words (strings).

Here's the code:

typedef vector<string> Pseudosentence;
#define W 20  // A Pseudosentence is made of W words
#define K 2   // A block is made of K Pseudosentences

class Block {
public:
vector<Pseudosentence> p;
multimap<string, int> Scoremap;
Block() {
    p.resize(2);
}
Block(Pseudosentence First, Pseudosentence Second){ 
    p.resize(2);
    p[0] = First;
    p[1] = Second;
}
void rankTerms(); // Calculates some ranking function
void setData(Pseudosentence First, Pseudosentence Second){
    p[0] = First; 
    p[1] = Second;
}
};
stringstream str(final); // Final contains the (preprocessed) text.
string t;
vector<Pseudosentence> V; // V[j][i]. Every V[j] is a pseudosentence. Every V[j][i] is a word (string).
vector<Block> Blocks;
vector<int> Score;
Pseudosentence Helper;
int i = 0;
int j = 0;
while (str) {
    str >> t;
    Helper.push_back(t);
    i++;
    //cout << Helper[i];
    if (i == W) {  // When I have a pseudosentence...
        V.push_back(Helper);
        j++; // This measures the j-th pseudosentence
        Helper.clear();
    }
    if (i == K*W) {
        V.push_back(Helper);
        j++; // This measures the j-th pseudosentence
        Helper.clear();
        //for (int q=0; q < V.size(); ++q) {
            //cout << "Cluster "<< q << ": \n";
            //for (int y=0; y < V[q].size(); ++y)       // This works
                //cout << y <<": "<< V[q][y] << endl;
        //}
        Block* Blbl = new Block;
        Blbl->setData(V[j-1], V[j]); // When I have K pseudosentences, I have a block.
        cout << "B = " << Blbl->p[0][5]<< endl;
        Blbl->rankterms(); // Assigning scores to words in a block
        Blocks.push_back(*Blbl);
        i = 0;
    }
}

The code compiles, but when I try to use the setData(a,b) method from Block, XCode takes me to stl_construct.h and tells me that he received a EXC_BAD_ACCESS signal.

The code to which I am taken is this:

/** @file stl_construct.h
*  This is an internal header file, included by other library headers.
*  You should not attempt to use it directly.
*/

#ifndef _STL_CONSTRUCT_H
#define _STL_CONSTRUCT_H 1

#include <bits/cpp_type_traits.h>
#include <new>

_GLIBCXX_BEGIN_NAMESPACE(std)

  /**
   * @if maint
   * Constructs an object in existing memory by invoking an allocated
   * object's constructor with an initializer.
   * @endif
   */
  template<typename _T1, typename _T2>
    inline void
    _Construct(_T1* __p, const _T2& __value)
    {
      // _GLIBCXX_RESOLVE_LIB_DEFECTS
      // 402. wrong new expression in [some_]allocator::construct
      ::new(static_cast<void*>(__p)) _T1(__value);
    }

(The actual line that XCode highlights is the ::new(static_cast<void*>(__p)) _T1(__value); so I thought it was due to the new operator, but in fact the debugger showed me that I can use a new Block; What I can't do is a new Block(a,b) (with the parameter constructor) or setting data... I find this awkward, because every documentation says that the = operator has been overloaded for vectors, so it should be no problem... Sorry again for the stupid question, but I can't find it. :-(

Upvotes: 4

Views: 635

Answers (1)

SoapBox
SoapBox

Reputation: 20609

Every time you add an element to V you also increment j. This means that j will always equal the length of V.

That means that the below line will always result in access 1 past the end of V.

Blbl->setData(V[j-1], V[j]);

Using that value later (when it it part of a Block's p vector will result in all manner of potential problems. This is likely the source of your issue.

Also, you have a memory leak (you newed but didn't delete). Use a scoped_ptr here, or just create the value on the stack. There doesn't seem to be a reason to allocate it on the heap.

Upvotes: 2

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