这个问题应该很简单,也许很愚蠢,但我找不到问题。STL Vectors and the new operator
基本上,我必须用自然语言解析一些句子。我需要实现一个操纵“块”的简单算法。一个Block由2个假词构成,由20个字(字符串)组成。
下面的代码:
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;
}
}
代码编译,但是当我尝试从模块使用setData(a,b)
方法的XCode带我到stl_construct.h
并告诉我,他收到了EXC_BAD_ACCESS
信号。
到我采取的代码是这样的:
/** @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);
}
(即XCode中突出了实际的线是::new(static_cast<void*>(__p)) _T1(__value);
所以我认为这是由于新的运营商,但事实上,调试器显示我,我可以使用一个新的块;我不能做的是一个新的Block(a,b)
(带参数构造函数)或设置数据...我觉得这很尴尬,因为每个文档都说=
运算符已经为矢量重载,所以它应该没有问题...对不起,我再也找不到它了。:-(
这怎么可能编译?我找不到类声明的结尾,也找不到SetData。 – Lou
你可以格式化你的代码吗?不要使用标签,只能使用空格。缩进和空白将不胜感激。 –
对不起,我没有粘贴setData方法,我以为我有。我现在格式化它 – Tex