C++ For 循环上的分段错误
在索引 19 的最后一个值之前，一切都工作正常。事实上，所有值都被打印出来，还有什么没有。一旦它打印出最终值&索引，它分段故障。我假设这是因为它正在尝试访问第 20 个值。我该如何防止这种情况发生？
主文件代码：
int index = 0;
while (index < list.length())
{
cout << list.getNextItem(index) << " " << index << "\n";
index++;
}
标题代码：
template <class Type>
Type doublyLinkedList<Type>::getNextItem(const Type& val) const
{
nodeType<Type> *current; //pointer to traverse the list
current = first; //set current to point to the first node
for (index=0; index < val; index++)
{
if (current != NULL)
{
current = current->next;
}
}
return current->info;
}//end getNextItem
Everything works fine right up until the last value of index 19. In fact, all values are printed and what not. Once it prints the final value & index, it seg faults. I am assuming this is because it is trying to access the 20th value. How would I prevent this from occurring?
MAIN FILE CODE:
int index = 0;
while (index < list.length())
{
cout << list.getNextItem(index) << " " << index << "\n";
index++;
}
HEADER CODE:
template <class Type>
Type doublyLinkedList<Type>::getNextItem(const Type& val) const
{
nodeType<Type> *current; //pointer to traverse the list
current = first; //set current to point to the first node
for (index=0; index < val; index++)
{
if (current != NULL)
{
current = current->next;
}
}
return current->info;
}//end getNextItem
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评论（2）
您将
当前分配给当前->下一个。当它为null那么，当current为 ... null 时，您尝试返回current->info。至少这是我的怀疑；您发布的代码不完整，不可能给您具体的答案……但这看起来确实是罪魁祸首。
You assign
currenttocurrent->next. When it'snullthen you to try to returncurrent->infowhencurrentis ... null.At least that's my suspicion; the code you post is incomplete and it's impossible to give you a concrete answer ... but that certainly looks like a culprit.
您的
current->info不在空检查范围内。当current为null时，无法访问其指针并导致seg错误Your
current->infois outside null checking. Whencurrentis null, you can't access its pointer and cause seg fault