blocks|key|1225643|text|为什么不直接使用库例程wcstombs。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1225644|entityMap|0|LINK|mutability|MUTABLE|url|http://www.thinkage.ca/english/gcos/expl/c/lib/wcstom.html^0|B|8|B|8|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Why not just use a library routine <a href="http://www.thinkage.ca/english/gcos/expl/c/lib/wcstom.html" rel="noreferrer"><code>wcstombs</code></a>.

blocks|key|1011193|text|assert用于确保在调试模式下某些内容是真的，而不会在发布版本中产生任何影响。最好使用if语句，并为超出范围的字符制定替代计划，除非获取超出范围的字符的唯一方法是通过程序错误。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1011194|此外，根据字符编码的不同，您可能会发现从0x80到0xff的Unicode字符与其char版本之间的差异。|1011195|entityMap^0|0|6|18|2|0|15|4|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

<code>assert</code> is for ensuring that something is true in a debug mode, without it having any effect in a release build. Better to use an <code>if</code> statement and have an alternate plan for characters that are outside the range, unless the only way to get characters outside the range is through a program bug.

Also, depending on your character encoding, you might find a difference between the Unicode characters 0x80 through 0xff and their <code>char</code> version.

blocks|key|1225775|text|从技术上讲，'char‘可以与'signed+char’或'unsigned+char‘具有相同的范围。对于无符号字符，您的范围是正确的；对于有符号字符，理论上，您的条件是错误的。在实践中，很少有编译器会反对--结果也是一样的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1225776|吹毛求疵：assert中的最后一个&&是一个语法错误。|1225777|断言是否合适取决于您是否能够承受代码到达客户时发生崩溃的后果，以及如果违反了断言条件但断言没有编译到代码中，您可以或应该做什么。对于调试工作，它看起来很好，但您可能还需要在它之后进行活动测试，以便进行运行时检查。|1225778|entityMap^0|7|4|G|B|U|D|0|5|6|H|2|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|X|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|K|$]]

Technically, '<code>char</code>' could have the same range as either '<code>signed char</code>' or '<code>unsigned char</code>'. For the unsigned characters, your range is correct; theoretically, for signed characters, your condition is wrong. In practice, very few compilers will object - and the result will be the same.

Nitpick: the last <code>&amp;&amp;</code> in the <code>assert</code> is a syntax error.

Whether the assertion is appropriate depends on whether you can afford to crash when the code gets to the customer, and what you could or should do if the assertion condition is violated but the assertion is not compiled into the code. For debug work, it seems fine, but you might want an active test after it for run-time checking too.

blocks|key|1225840|text|一般来说，不是。当然是int(wchar_t(255))+==+int(char(255))，但这只是意味着它们有相同的int值。它们可能代表不同的字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1225841|你甚至会在大多数Windows个人电脑中看到这样的差异。例如，在Windows代码页1250上，char(0xFF)与wchar_t(0x02D9)+(上面的点)是相同的字符，而不是wchar_t(0x00FF)+(带分隔符的小写y)。|1225842|注意，它甚至不适用于ASCII码范围，因为C%2B%2B甚至不需要ASCII码。特别是在IBM系统上，您可能会看到'A'+!=+65|1225843|entityMap^0|B|Z|0|1C|A|1N|F|2J|F|0|1H|9|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|K|$]]

In general, no. <code>int(wchar_t(255)) == int(char(255))</code> of course, but that just means they have the same int value. They may not represent the same characters.

You would see such a discrepancy in the majority of Windows PCs, even. For instance, on Windows Code page 1250, <code>char(0xFF)</code> is the same character as <code>wchar_t(0x02D9)</code> (dot above), not <code>wchar_t(0x00FF)</code> (small y with diaeresis).

Note that it does not even hold for the ASCII range, as C++ doesn't even require ASCII. On IBM systems in particular you may see that <code>'A' != 65</code>

blocks|key|1225881|text|还可以转换wchar_t+-->+wstring+-->+string+-->+char|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1225882|wchar_t+wide;
wstring+wstrValue;
wstrValue[0]+=+wide

string+strValue;
strValue.assign(wstrValue.begin(),+wstrValue.end());++//+convert+wstring+to+string

char+char_value+=+strValue[0];|code-block|syntax|javascript|1225883|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

one could also convert wchar_t --> wstring --> string --> char 

<pre><code>wchar_t wide;
wstring wstrValue;
wstrValue[0] = wide

string strValue;
strValue.assign(wstrValue.begin(), wstrValue.end()); // convert wstring to string

char char_value = strValue[0];
</code></pre>

blocks|key|1011382|text|我写了一个简短的函数，将一个wchar_t数组打包成一个字符数组。不在ANSI代码页(0-127)上的字符将替换为'?‘字符，并且它可以正确处理代理项对。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1011383|size_t+to_narrow(const+wchar_t+*+src,+char+*+dest,+size_t+dest_len){
++size_t+i;
++wchar_t+code;

++i+=+0;

++while+(src[i]+!=+'\0'+&&+i+<+(dest_len+-+1)){
++++code+=+src[i];
++++if+(code+<+128)
++++++dest[i]+=+char(code);
++++else{
++++++dest[i]+=+'?';
++++++if+(code+>=+0xD800+&&+code+<=+0xD8FF)
++++++++//+lead+surrogate,+skip+the+next+code+unit,+which+is+the+trail
++++++++i%2B%2B;
++++}
++++i%2B%2B;
++}

++dest[i]+=+'\0';

++return+i+-+1;

}|code-block|syntax|javascript|1011384|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A short function I wrote a while back to pack a wchar_t array into a char array. Characters that aren't on the ANSI code page (0-127) are replaced by '?' characters, and it handles surrogate pairs correctly.

<pre><code>size_t to_narrow(const wchar_t * src, char * dest, size_t dest_len){
 size_t i;
 wchar_t code;

 i = 0;

 while (src[i] != '\0' &amp;&amp; i &lt; (dest_len - 1)){
 code = src[i];
 if (code &lt; 128)
 dest[i] = char(code);
 else{
 dest[i] = '?';
 if (code &gt;= 0xD800 &amp;&amp; code &lt;= 0xD8FF)
 // lead surrogate, skip the next code unit, which is the trail
 i++;
 }
 i++;
 }

 dest[i] = '\0';

 return i - 1;

}
</code></pre>

blocks|key|1225963|text|这是另一种方法，记住对结果使用free()。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1225964|char*+wchar_to_char(const+wchar_t*+pwchar)
{
++++//+get+the+number+of+characters+in+the+string.
++++int+currentCharIndex+=+0;
++++char+currentChar+=+pwchar[currentCharIndex];

++++while+(currentChar+!=+'\0')
++++{
++++++++currentCharIndex%2B%2B;
++++++++currentChar+=+pwchar[currentCharIndex];
++++}

++++const+int+charCount+=+currentCharIndex+%2B+1;

++++//+allocate+a+new+block+of+memory+size+char+(1+byte)+instead+of+wide+char+(2+bytes)
++++char*+filePathC+=+(char*)malloc(sizeof(char)+*+charCount);

++++for+(int+i+=+0;+i+<+charCount;+i%2B%2B)
++++{
++++++++//+convert+to+char+(1+byte)
++++++++char+character+=+pwchar[i];

++++++++*filePathC+=+character;

++++++++filePathC+%2B=+sizeof(char);

++++}
++++filePathC+%2B=+'\0';

++++filePathC+-=+(sizeof(char)+*+charCount);

++++return+filePathC;
}|code-block|syntax|javascript|1225965|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's another way of doing it, remember to use free() on the result.

<pre><code>char* wchar_to_char(const wchar_t* pwchar)
{
 // get the number of characters in the string.
 int currentCharIndex = 0;
 char currentChar = pwchar[currentCharIndex];

 while (currentChar != '\0')
 {
 currentCharIndex++;
 currentChar = pwchar[currentCharIndex];
 }

 const int charCount = currentCharIndex + 1;

 // allocate a new block of memory size char (1 byte) instead of wide char (2 bytes)
 char* filePathC = (char*)malloc(sizeof(char) * charCount);

 for (int i = 0; i &lt; charCount; i++)
 {
 // convert to char (1 byte)
 char character = pwchar[i];

 *filePathC = character;

 filePathC += sizeof(char);

 }
 filePathC += '\0';

 filePathC -= (sizeof(char) * charCount);

 return filePathC;
}
</code></pre>

I was wondering is it safe to do so?

<pre><code>wchar_t wide = /* something */;
assert(wide &gt;= 0 &amp;&amp; wide &lt; 256 &amp;&amp;);
char myChar = static_cast&lt;char&gt;(wide);
</code></pre>

If I am pretty sure the wide char will fall within ASCII range.

Convert wchar_t to char

编译器

Windows

我想知道这样做安全吗？wchar_t wide = /* something */;assert(wide >= 0 && wide < 256 &&);char myChar = static_cast<char>(wide);如果我非常确定宽字符将落在ASCII范围内。

问将wchar_t转换为字符
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