对向量使用push_back< vector< string> > [英] Using push_back on a vector<vector<string> >

问题描述

我有点尴尬，这样一个简单的问题阻碍了我，但经过几个小时无果的谷歌，我仍然卡住。

为了简化我的问题，此第二行崩溃：

  vector< vector< string> > sorted_words; 
 sorted_words [0] .push_back（hello）; 
  

不应 sorted_words [0] 一个空的向量，我可以合法push_back到？

解决方案

不。您有一个向量的向量。你没有向第二行添加任何东西，所以 sorted_words [0] 是 sorted_words ，还不存在。

您尝试将hello推送到一个空的向量。

空指针解引用

我会问你真的想要一个向量

如果你想要一个字符串的向量，请使用：

<$> p $ p> vector< string> sorted_words;
sorted_words.push_back（hello）;


如果你确实需要一个向量（字符串）的向量，

 矢量< string> first_vector; 
 first_vector.push_back（hello）; 
 sorted_words.push_back（first_vector）; 
  

I'm somewhat embarrassed that such a simple problem has stymied me, but after a few hours of fruitless googling, I'm still stuck.

To simplify my problem, the 2nd line of this crashes:

  vector<vector<string> > sorted_words;
  sorted_words[0].push_back("hello");

Shouldn't sorted_words[0] represent an empty vector that I can legally push_back onto?

解决方案

No. You have a vector of vectors. You haven't added anything to either vector by line 2, so sorted_words[0], the first element in sorted_words, doesn't exist yet.

You're trying to push "hello" into a null vector.

Null pointer dereference!

I would ask "do you really want a vector of vectors, or just a vector of strings"?

If you want a vector of strings, then use:

vector<string> sorted_words;
sorted_words.push_back("hello");

If you really do want a vector of vectors (of strings), then use:

vector<string> first_vector;
first_vector.push_back("hello");
sorted_words.push_back(first_vector);

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