为什么下面的C++代码只打印第一个字符？

发布于 2024-10-08 11:48:26 字数 390 浏览 6 评论 0原文

我正在尝试将 char 字符串转换为 wchar 字符串。

更详细地说：我试图首先将 char[] 转换为 wchar[]，然后将“1”附加到该字符串并打印它。

char src[256] = "c:\\user";

wchar_t temp_src[256];
mbtowc(temp_src, src, 256);

wchar_t path[256];

StringCbPrintf(path, 256, _T("%s 1"), temp_src);
wcout << path;

但它只打印 c

这是从 char 转换为 wchar 的正确方法吗？从那时起我就知道了另一种方法。但我想知道为什么上面的代码会这样工作？

原文

I am trying to convert a char string to a wchar string.

In more detail: I am trying to convert a char[] to a wchar[] first and then append " 1" to that string and the print it.

char src[256] = "c:\\user";

wchar_t temp_src[256];
mbtowc(temp_src, src, 256);

wchar_t path[256];

StringCbPrintf(path, 256, _T("%s 1"), temp_src);
wcout << path;

But it prints just c

Is this the right way to convert from char to wchar? I have come to know of another way since. But I'd like to know why the above code works the way it does?

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从﹋此江山别 2024-10-15 11:48:26

mbtowc 仅转换单个字符。您是否打算使用mbstowcs？

通常您会调用该函数两次；第一个获取所需的缓冲区大小，第二个实际转换它：

#include <cstdlib> // for mbstowcs

const char* mbs = "c:\\user";
size_t requiredSize = ::mbstowcs(NULL, mbs, 0);
wchar_t* wcs = new wchar_t[requiredSize + 1];
if(::mbstowcs(wcs, mbs, requiredSize + 1) != (size_t)(-1))
{
    // Do what's needed with the wcs string
}
delete[] wcs;

如果您更愿意使用 mbstowcs_s（因为弃用警告），然后执行以下操作：

#include <cstdlib> // also for mbstowcs_s

const char* mbs = "c:\\user";
size_t requiredSize = 0;
::mbstowcs_s(&requiredSize, NULL, 0, mbs, 0);
wchar_t* wcs = new wchar_t[requiredSize + 1];
::mbstowcs_s(&requiredSize, wcs, requiredSize + 1, mbs, requiredSize);
if(requiredSize != 0)
{
    // Do what's needed with the wcs string
}
delete[] wcs;

确保通过 setlocale() 或使用 mbstowcs() 版本（例如 mbstowcs_l()或采用区域设置参数的 mbstowcs_s_l())。

mbtowc converts only a single character. Did you mean to use mbstowcs?

Typically you call this function twice; the first to obtain the required buffer size, and the second to actually convert it:

#include <cstdlib> // for mbstowcs

const char* mbs = "c:\\user";
size_t requiredSize = ::mbstowcs(NULL, mbs, 0);
wchar_t* wcs = new wchar_t[requiredSize + 1];
if(::mbstowcs(wcs, mbs, requiredSize + 1) != (size_t)(-1))
{
    // Do what's needed with the wcs string
}
delete[] wcs;

If you rather use mbstowcs_s (because of deprecation warnings), then do this:

#include <cstdlib> // also for mbstowcs_s

const char* mbs = "c:\\user";
size_t requiredSize = 0;
::mbstowcs_s(&requiredSize, NULL, 0, mbs, 0);
wchar_t* wcs = new wchar_t[requiredSize + 1];
::mbstowcs_s(&requiredSize, wcs, requiredSize + 1, mbs, requiredSize);
if(requiredSize != 0)
{
    // Do what's needed with the wcs string
}
delete[] wcs;

Make sure you take care of locale issues via setlocale() or using the versions of mbstowcs() (such as mbstowcs_l() or mbstowcs_s_l()) that takes a locale argument.

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╰◇生如夏花灿烂 2024-10-15 11:48:26

为什么你使用C代码，为什么不以更可移植的方式编写它，例如我在这里要做的是使用STL！

std::string  src = std::string("C:\\user") +
                   std::string(" 1");
std::wstring dne = std::wstring(src.begin(), src.end());

wcout << dne;

就这么简单，很容易：D

why are you using C code, and why not write it in a more portable way, for example what I would do here is use the STL!

std::string  src = std::string("C:\\user") +
                   std::string(" 1");
std::wstring dne = std::wstring(src.begin(), src.end());

wcout << dne;

it's so simple it's easy :D

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