给定一个字符串数组,我正在编写一个类,根据它们的长度将它们分成不同的组,即相同长度的组转到同一组。组的数量和每组的大小是未知的。
我的想法是:
我使用私有数据成员std::vector<std::vector<std::int> > m_groups
,意图是外部向量维护组,内部向量跟踪属于一个组的字符串的所有索引。
问题是,一旦我将字符串推入向量,我的一些数据成员就会被破坏。有谁可以看看?
以下是简化代码:
class A {
public:
A(std::string words[], int num, int c1[], int m, int c2[], int n);
~A();
void print_state();
private:
int *m_var;
int m_Nvar;
std::vector<std::vector<std::int> > m_doms;
std::vector<std::vector<std::int> > m_groups;
std::vector<std::string> > m_words;
int *m_cst1;
int *m_cst2;
int m_Ncst;
};
在构造函数中: A :: print_cst2() { for(int c = 0; c&lt; m_Ncst; c ++) { printf(“%d”,m_cst2 [4 * c]); printf(“%d”,m_cst2 [4 * c + 1]); printf(“%d”,m_cst2 [4 * c + 2]); printf(“%d”,m_cst2 [4 * c + 3]); } }
A::A(std::string words[], int num,
int c1[], int m, int c2[], int n) {
...
m_cst1 = new int[m/2];
m_cst2 = new int[n/4];
m_Ncst = n/4;
m_Nvar = m/2;
for (int i = 0; i < n; i+=4)
{
m_cst2[i] = c2[i];
m_cst2[i+1] = c2[i+1];
m_cst2[i+2] = c2[i+2];
m_cst2[i+3] = c2[i+3];
}
print_cst2(); // (1) we print the m_cst2
// we are only interested, the words of length smaller than m_max_len
// put m_max_len number of empty vectors (groups) in the group vector
for (int i = 0; i < m_max_len; i++)
{
m_groups.push_back(std::vector<int>());
}
print_cst2(); // (2) we print the m_cst2 again
// go through every words and copy words of interest to m_words
// push the index of the word to the group it belongs to (by its length)
for (int i = 0, k = 0; i < num; i++, k++)
{
int len = words[i].length();
if (len > m_max_len)
continue;
m_words.push_back(words[i]);
m_groups[len].push_back(k);
}
// you can ignore this part: link the group to another structure
for (int i = 0; i < m_Nvar; i++)
{
m_doms.push_back(m_groups[m_cst1[i]]);
}
...
}
...
我编译了代码并运行。数组m_cst2
末尾的数据已损坏。这似乎与std::vector
的使用有关。 User comingstorm provided an interesting clue:外部std::vector
在其堆分配中存储一个固定大小的std::vector<int>
数据结构数组。这是答案吗?不过不确定。所以,发布这个并征求建议。
PS:如果您有更好的想法来完成这项任务,请告诉我......如果您需要更多信息,请发布。
我感谢你的时间。
答案 0 :(得分:1)
以下一些改进。特别是你有一个off by 1索引错误,并且k
的增量不正确:
CwordSolver::CwordSolver(std::string words[], int num,
int c1[], int m, int c2[], int n) {
...
// we are only interested, the words of length smaller than m_max_len
m_groups.resize(0);
m_groups.resize(m_max_len, std::vector<int>());
// go through every words and copy words of interest to m_words
// push the index of the word to the group it belongs to (by its length)
int k = 0; // the index in m_words
for (int i = 0; i < num; i++)
{
int len = words[i].length();
if (len >= m_max_len)
continue;
m_words.push_back(words[i]);
m_groups[len].push_back(k);
k++;
}
...
}
答案 1 :(得分:0)
如果您发布的代码是您正在使用的实际代码,则会损坏内存。
假设n
是4.
m_cst2 = new int[n/4]; // so you have room for 1 item
m_Ncst = n/4;
m_Nvar = m/2;
for (int i = 0; i < n; i+=4)
{
m_cst2[i] = c2[i]; // valid
m_cst2[i+1] = c2[i+1]; // memory overwrite
m_cst2[i+2] = c2[i+2]; // memory overwrite
m_cst2[i+3] = c2[i+3]; // memory overwrite
}
没有必要走得更远。 n
的值为4,当您写入m_cst2
等时,显然您将在m_cst2[1], m_cst2[2]
中超出范围,因为唯一有效的条目是{{1} }}
所以你的内存损坏可能与m_cst2[0]
无关(很难弄乱对象的向量),而且很可能与上面的代码有关。
此外,您应该在代码中使用std::vector
而不是std::vector
。如果您将new[]/delete[]
用于一件事,为什么不随时随地利用它?