在Windows上使用std :: copy复制宽字符串

Question

Let's say I have a wide string like this

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\n",&c[0] );

Second

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\n",&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

Why so much hotchpotch?

I wonder why you're not doing this:

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

Simple, and concise!

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.

---

If you use std::copy , for example when you cannot use vector constructor for some reason, then the idiomatic std::copy is this:

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:

If you're working with C library, and APIs which works with c-string, then I would suggest you to use std::wstring (or std::string ). I don't see much point in using std::vector , for it adds confusion to the code. If you use vector, you have to append a null character at the end, so as to make it work:

  c.push_back('\0'); //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

当您要求一个std::vector<wchar_t> c(size)它会分配足够的空间来容纳大小为wchar_t项目-您要求的大小不是以字节为单位，而是以wchar_t项的形式。

Answer 3

First: You don't actually need an ending null character, when you have a size-aware container.

Secondly, this constructor for std::vector<T> takes the number of T elements, not bytes. So +1 means one extra wchar_t(0) .

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