Convert std::wstring to WCHAR array

Question

I have been searching the internet for days about this question. I have made a win32 project in which I want to convert a wstring to a WCHAR array

Please give an example If you find out any mistakes please give an example

wstring timeNow = L"Hello";
WCHAR timeWchar[6] = {(WCHAR)timeNow.c_str()}; // Not Working

Instead of the text I see only a square when I run my program

Does this answer your question? [How to convert wstring to wchar\_t\*? C++](https://stackoverflow.com/questions/44985451/how-to-convert-wstring-to-wchar-t-c) — Simon Mourier, Aug 07 '21 at 09:13
Does it really matter that the WCHAR array is 6 large? Since you are creating an array, you need to copy over the values into it. You could also use `const WCHAR *timeWchar = timeNow.c_str();` that way, you can also use functions that take `const WCHAR [ ]` arguments without the need to copy. — JVApen, Aug 07 '21 at 09:20
Your code fails because the cast tells a compiler a lie. A pointer is not a WCHAR. That's why the compiler objected, and why the cast allowed the code to compile but not do anything useful. A question for me is what you are doing with this array? As JVApen says perhaps you don't need the array at all depending on what the code that uses it is actually doing. We can't see that unfortunately. — David Heffernan, Aug 07 '21 at 10:05
I left you an answer below showing possibilities of initializing an array from a string. But it occurred to me, just like the comments above that I didn't originally read, that copying a string into an array may not be needed at all. There's a lot of great `std::` classes in C++ that would probably be better than using a string copy into an array for the task at hand. Can you open up more about what you are really trying to do? I'll amend my answer if I have more details from you. — selbie, Aug 07 '21 at 17:13

score 0 · Answer 1 · answered Aug 07 '21 at 08:50

0

I assume that by WCHAR, you mean wchar_t.

You can loop over the array and assign the elements.

Or if you don't feel like writing the loop yourself, you can use an algorithm from the standard library. Example:

assert(timeNow.size() < 6);
wchar_t timeWchar[6] {};
std::ranges::copy(timeNow, timeWchar);

answered Aug 07 '21 at 08:50

eerorika

232,697
12
197
326

re the first sentence of this answer, isn't WCHAR just an alias to wchar_t? – David Heffernan Aug 07 '21 at 10:03
@DavidHeffernan That's what I've assumed. But technically it is a non-reserved identifier, so OP could have defined it as any kind of entity. – eerorika Aug 07 '21 at 10:05
but why `< 6`? should be `< 7` – Алексей Неудачин Aug 07 '21 at 10:34
@АлексейНеудачин `but why < 6?` Because a longer string won't fit inside the array without being truncated (assuming null termination). Edit: `should be < 7` then there would be no room for the null terminator. – eerorika Aug 07 '21 at 10:36
`L'\0'` doesn't go to size. if it is (you think it's there) in the end – Алексей Неудачин Aug 07 '21 at 10:36
you have array on out , not a string. And you can have `L'\0'` in the middle with np in both wstr and array here – Алексей Неудачин Aug 07 '21 at 10:42
`std::string t = {'a','b'};` – Алексей Неудачин Aug 07 '21 at 10:51

score 0 · Answer 2 · answered Aug 07 '21 at 10:41

0

You can't really initialize an array with a pointer.

This is closer to what you want without making a copy of the string into an array. just use a pointer to reference the chars.

wstring timeNow = L"Hello";
const WCHAR* timeWchar = timeNow.c_str();

99% of the time, the above works for whatever you need to do assuming you don't need to modify the chars.

If really need to make a make a copy of the characters into a different array, such as when you need to manipulate the string, this will do in Windows just fine - assuming you have a fixed size array that is big enough to receive the copy.

wstring timeNow = L"Hello";
WCHAR timeWchar[6];
StringCchCopyW(timeWchar, ARRAYSIZE(timeWchar), timeNow.c_str());

If you don't know the length ahead of time, then you'll need to allocate it before making the copy:

size_t allocSize = timeNow.size() + 1;
WCHAR timeWchar = new WCHAR[allocSize];
StringCchCopyW(timeWchar, allocSize, timeNow.c_str());
/* don't forget to `delete [] timeWchar` when you are done */

But then again why mess with new and delete when you can just let C++ do the work for you. Make a copy of the string and then use a pointer to reference the characters in the copy.

wstring timeNowCopy = timeNow;
const WCHAR* timeWchar = timeNowCopy.c_str(); // timeWchar points to the array copied into timeNowCopy.

If you can maintain the lifetime of timeNowCopy and timeWchar on the stack together, then you don't need to explicitly new or delete anything.

answered Aug 07 '21 at 10:41

selbie

100,020
15
103
173

if someone goes with array for a reason `StringCchCopyW` would not work anyway. Because there're(or may be) `\0`'s – Алексей Неудачин Aug 07 '21 at 11:00
i'll go with `memcpy( .., wstr.data(), wstr.size()*2)` – Алексей Неудачин Aug 07 '21 at 11:12
@АлексейНеудачин - you would be correct to suggest `memcpy`, but it would be an exceptional case where the user needs to preserve null chars from a string. I also suspect that the OP only thinks he needs an array. Whatever he thinks he needs an array of `WCHAR` for, he can easily accomplish with `std::wstring`. – selbie Aug 07 '21 at 17:07
yeah he is. it's if you really need array. – Алексей Неудачин Aug 07 '21 at 17:11
I suspect if the OP is new to C++ and isn't familiar with converting a string to an array, he probably doesn't need to be doing that anyway. I've left the OP a comment about explaining what he's really trying to accomplish – selbie Aug 07 '21 at 17:15
i mean because he doesn't need it he also doesn't need a `WCHAR timeWchar = new WCHAR[allocSize];`. And if someone need this then use of `StringCchCopyW` is wrong . – Алексей Неудачин Aug 07 '21 at 17:20

Алексей Неудачин · Answer 3 · 2021-08-07T10:27:21.867

-2

just do its .data() .

WCHAR timeWchar[6] = ...

wont work anyway

in which I want to convert a wstring to a WCHAR array

if you mean wchar_t [6] type - you just do not need this. you don't have a situation where you can go with wchar_t [6] but can't with wstr.data() .

edited Aug 07 '21 at 10:27

answered Aug 07 '21 at 10:08

Алексей Неудачин

405
4
13

1

*"you don't have a situation where you can go with `wchar_t [6]` but can't with `wstr.data()`" - That's incorrect. `wstr.data()` loses the size information, that's still captured in the array. – IInspectable Aug 07 '21 at 11:13

Convert std::wstring to WCHAR array

3 Answers3