Question

让我说一下,我有这么多的经历。

std::wstring s(L"Stack Over Flow");

that I want to copy into a vector of wide characters std::vector<wchar_t> (of a null terminated string ) using std::copy
Both the following approaches seem to work
First

std::vector<wchar_t> c( s.size() + 1 );
std::copy(s.c_str(), s.c_str() + s.length() + 1, c.begin());
wprintf( L"%s
",&c[0] );

二、导言

std::vector<wchar_t> d( s.size() + sizeof(wchar_t));
std::copy(s.c_str(), s.c_str() + s.length() + sizeof(wchar_t), d.begin());
wprintf( L"%s
",&d[0] );

But I am thinking the first should fail because I am not allocating the correct size for the ending null character. What did I miss?
Edit
No, I don t want to to this

std::vector<wchar_t> c(s.begin(), s.end()); <br>

because I am resizing an existing buffer
Edit
By changing the use case (print using the C like API ) I can see that the second version is probably the correct one.
First prints Stack Over Flow????????
Second prints Stack Over Flow as expected

Answer 1

为什么如此多的热点?

我想知道,为什么你不这样做:

std::vector<wchar_t> c(s.begin(), s.end());

简明扼要!

As for your code, both versions are wrong. Both are wrong because the second argument is going beyond the end, as it should be only s.c_str() + s.length(). That is,

2nd arg ---> s.c_str() + s.length() //correct

2nd arg ---> s.c_str() + s.length() + 1 //wrong - first version
2nd arg ---> s.c_str() + s.length() + sizeof(wchar_t) //wrong - second version

If you follow simple code, idiomatic code, you can avoid much of error.

如果您使用<代码>std:copy,例如,如果您因某种原因不能使用病媒构造,那么辅助器<代码>则:copy:

void f(std::vector<wchar_t> & c)
{
  //you cannot use constructor version now, so use std::copy as
  std::copy(s.begin(), s.end(), std::back_inserter(c));
}

EDIT:

如果你重新与C间图书馆和有方位的APICS合作,那么我建议你使用<代码>std:wstring <>code>(或std:string)。我在使用<代码>std:vector方面看不出什么点,因为它给守则增加了混乱。如果你使用病媒,你最终必须证明无效,以便发挥作用:

  c.push_back(  ); //do this after copying!

But I still don t see any point in doing that, as you re trying to treat vector like a string. If you need string, why not use std::wstring and it s std::wstring::c_str() to get the c-string.

Answer 2

When you ask for a std::vector<wchar_t> c(size) it allocates enough space to contain size wchar_t items - the size you ask for is not in bytes, but is in wchar_t items.

Answer 3

首先:当你拥有一个大小设备集装箱时,你实际上并不需要终止无效。

第二,<代码>的这个构件: T> 系指T元素的数量,而不是tes。 http://code>+1。

[edit] With the edited question, the easiest solution seems c.assign(s.begin(), s.end()).

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