blocks|key|4403295|text|为什么all_vowels()要分配内存？更有趣的是，为什么不free()它呢？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4403296|我非常确定all_vowels()不需要分配任何内存，而且可以比你拥有的简单一点。|4403297|此外，在尝试读取文件之前，您不能使用feof()。去掉它，循环直到fgets()返回NULL。|4403298|我可能会编写一个辅助函数int+is_vowel(char+c);来使代码更清晰，然后在all_vowels()中像这样解决这个问题|4403299|vowels+=+"aeiou"
for+each+character+x+in+the+string+to+check:
++if+is_vowel(x):
++++if+vowels+starts+with+x:
++++++let+vowels+=+vowels[1:]
++++else
++++++return+false
return+true+if+vowels+is+empty|code-block|syntax|javascript|4403300|entityMap^0|3|C|V|6|0|5|C|0|I|6|X|7|16|4|0|C|L|18|C|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|L|3|M|5|N|7|1C|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|R|$]]

Why is <code>all_vowels()</code> allocating memory? And, even more interesting, why doesn't it <code>free()</code> it?

I'm pretty sure <code>all_vowels()</code> doesn't have to allocate any memory, and can be a bit simpler than what you have.

Also, you can´t use <code>feof()</code> before trying to read from the file. Remove that, just loop until <code>fgets()</code> returns <code>NULL</code>.

I would probably write a helper function <code>int is_vowel(char c);</code> to make the code clearer, and then attack the problem like so in <code>all_vowels()</code>:

<pre><code>vowels = "aeiou"
for each character x in the string to check:
 if is_vowel(x):
 if vowels starts with x:
 let vowels = vowels[1:]
 else
 return false
return true if vowels is empty
</code></pre>

blocks|key|2678295|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2678296|，但它在某些地方有缺陷...|blockquote|2678297|2678298|2678299|这里可能有错误吗？|2678300|if+(+*vowel++++++==+'a'+&&
++++(*vowel+%2B+1)+==+'e'+&&
++++(*vowel+%2B+2)+==+'i'+&&
++++(*vowel+%2B+3)+==+'o'+&&
++++(*vowel+%2B+4)+==+'e'+)++++++
//++++++++++++++++++%5E%5E%5E+'u'?|code-block|syntax|javascript|2678301|还可能有其他错误。我还没有检查你所有的代码。|2678302|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|Q|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|R|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|T|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|U|7|@]|8|@]|9|$]]|$1|H|3|I|4|J|6|V|7|@]|8|@]|9|$K|L]]|$1|M|3|N|4|5|6|W|7|@]|8|@]|9|$]]|$1|O|3|-4|4|5|6|X|7|@]|8|@]|9|$]]]|P|$]]

<blockquote>
 but its flawed somewhere...
</blockquote>

Could there be an error here?

<pre><code>if ( *vowel == 'a' &amp;&amp;
 (*vowel + 1) == 'e' &amp;&amp;
 (*vowel + 2) == 'i' &amp;&amp;
 (*vowel + 3) == 'o' &amp;&amp;
 (*vowel + 4) == 'e' ) 
// ^^^ 'u'?
</code></pre>

There may also be other errors. I haven't checked all your code.

blocks|key|2678361|text|这里有一个很大的缺陷：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2678362|char+*v+=+malloc(sizeof(char+*));|code-block|syntax|javascript|2678363|这只分配四个或八个字节(取决于您是在32位还是64位平台上)。我猜你想要的不止这些。|2678364|PS。在未来，你可能应该尝试更具体，而不是仅仅说“它有缺陷”。|2678365|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is a big flaw:

<pre><code>char *v = malloc(sizeof(char *));
</code></pre>

This only allocates four or eight bytes (depending on if you are on a 32 or 64 bit platform). I'm guessing you want a little more than that.

PS. In the future, you should probably try to be more specific instead of just saying that "it's flawed".

blocks|key|2394073|text|您正在访问v，就像它指向一个包含多个字符的位置一样，而实际上您只为一个char+*保留了空间(在32位机器上通常是4字节，在64位机器上通常是8字节)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2394074|char+*v+=+malloc(sizeof(char+*));|code-block|syntax|javascript|2394075|对于您试图在其中存储的内容，这可能是足够的，也可能是不够的；在您的示例中，是任何给定单词中的元音数量。|2394076|只要有可能，您应该避免动态分配；在您的情况下，您不需要它们，您可以声明一个固定大小的数组，而不是char*。|2394077|char+v[5];|2394078|除此之外，您必须检查您是否已经读取了5个元音，以便不超过数组大小；如果在5个元音之后遇到另一个元音，则无论如何都可以停止检查；当前遇到的元音必须是重复的元音，因此该单词将不合格。|2394079|寻址字符的方式也是一个问题。再次检查*做了什么:它立即取消对右侧表达式的引用。在你的例子中，它总是取消引用v，然后添加一些东西(这也是合法的，因为取消引用的结果是一个char)。因此，如果v指向的第一个字符是a，第二个字符是e，那么*v将生成'a'，(*v+%2B+1)将生成'b'，(*v+%2B2)将生成'c'，依此类推-你看，结果是字母a加上给定的数字；第一个字符后面的内容并不重要，因为您永远不会访问那里的值。要想用指针算法实现你想要的结果，你必须使用括号：*(v%2B1)+--也就是说，将指针v加1，然后解引用它。这将产生c字符串中从v开始的第二个字符，即'e'。请注意，使用上面声明的v，您可以简单地编写v[0]、v[1]、v[2]等来寻址每个字符。|2394080|除此之外，检查if条件中的最后一个比较，那里有一个'e‘而不是一个'u’。|2394081|顺便说一句，顺便说一句，还有一些需要考虑的事情:你的问题有一个根本不需要v/vowel变量的解决方案……只有一个整型变量！|2394082|entityMap^0|5|1|Z|6|0|0|0|0|0|0|I|1|38|2|3D|3|3H|8|3S|3|3W|7|46|3|6D|6|6U|1|7F|1|7Q|3|8F|4|8K|4|8P|4|0|0|10|1|12|5|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]|$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|23|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]]|D|@]|E|$]]|$1|Y|3|-4|5|6|7|29|8|@]|D|@]|E|$]]]|Z|$]]

You are accessing <code>v</code> as if it were pointing to a location holding a number of characters, when indeed you only reserve space for one single <code>char *</code> (usually 4 byte on a 32 bit machine and 8 byte on a 64 bit machine):

<pre><code>char *v = malloc(sizeof(char *));
</code></pre>

That might or might not be enough for what you are trying to store in it; in your case, the number of vowels, in any given word.

Whenever possible, you should avoid dynamic allocations; in your case, you don't need them, you can declare an array of fixed size instead of a char*:

<pre><code>char v[5];
</code></pre>

In addition to that, you have to check if you have read 5 vowels already, so that you don't exceed the array size; if, after 5 vowels, you encounter another one, you can stop the check anyway; the currently encountered one has to be a duplicate vowel, the word will therefore not qualify.

The way you address characters is also a problem. Check again what <code>*</code> does: it dereferences the expression immediately to the right. In your case, it will always dereference v, then add something to it (which is also legal, since the result of dereferencing is a char). So if the first character where v points to is an a, the second an e, then <code>*v</code> will yield <code>'a'</code>, <code>(*v + 1)</code> will yield <code>'b'</code>, <code>(*v +2)</code> will yield <code>'c'</code> and so on - you see, the result is an addition to the letter a by the given number; it doesn't matter at all what comes after the first character because you never access the value there. To achieve what you want with pointer arithmetic, you'd have to use parenthesis: <code>*(v+1)</code> - i.e., add 1 to the pointer <code>v</code>, then dereference it. This would yield the second character in the c string starting at <code>v</code>, i.e. <code>'e'</code>.
Note that with v declared as above you can simply write <code>v[0]</code>, <code>v[1]</code>, <code>v[2]</code> and so on to address each character.

Aside from that, check the last comparison in your if condition, you had an 'e' instead of an 'u' there.

By the way, as a side note, and something to think about: There is a solution to your problem which does not require the <code>v</code>/<code>vowel</code> variables at all... only a single integer variable!

blocks|key|4314287|text|好的..。所以我最终得到了正确的输出...任何提高效率的提示或技巧都将不胜感激。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4314288|int+all_vowels(+char+*s+)
{
const+unsigned+char+*p+=+(const+unsigned+char+*)s;

char+v[5];
int+i+=+0;
if(*p+==+'\0')+return+-1;

while(+*p+!=+'\0'+)
{
++++if(+(*p+==+'a'+%7C%7C+*p+=='e'+%7C%7C+*p+=='i'%7C%7C+*p+=='o'+%7C%7C+*p+=='u')+&&+(+i+<+5+)++++++++)
++++{
++++++++v[i]+=+*p;
++++++++i%2B%2B;
++++}
++++p%2B%2B;
}


if+(+(+v[0]+==+'a'++&&++v[1]+==+'e'++&&++v[2]+==+'i'++&&++v[3]+==+'o'++&&++v[4]+==+'u'+)+&&+(strlen(v)+==+5+))+
{+
++++return+1;+
}

return+-1;
}

int+main+(int+argc,+char+*argv[])
{
FILE+*file;
char+line[30];

if+((+file+=+fopen("/usr/share/dict/words",+"r")+)+==+NULL)+
{
++++fprintf(stderr,+"cannot+open+%25s\n",+"/usr/share/dict/words");
++++exit(1);
}+

while+(+fgets(line,+sizeof(line),+file)+)
{

++++if+(+all_vowels(line)+==+1+)
++++{
++++++++printf("%25s\n",+line);
++++}
}
fclose(file);
return+0;

}|code-block|syntax|javascript|4314289|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|$]]

Okay.. So I finally got the right output... Any hints or tips for greater efficiency would be much appreciated.

<pre><code>int all_vowels( char *s )
{
const unsigned char *p = (const unsigned char *)s;

char v[5];
int i = 0;
if(*p == '\0') return -1;

while( *p != '\0' )
{
 if( (*p == 'a' || *p =='e' || *p =='i'|| *p =='o' || *p =='u') &amp;&amp; ( i &lt; 5 ) )
 {
 v[i] = *p;
 i++;
 }
 p++;
}


if ( ( v[0] == 'a' &amp;&amp; v[1] == 'e' &amp;&amp; v[2] == 'i' &amp;&amp; v[3] == 'o' &amp;&amp; v[4] == 'u' ) &amp;&amp; (strlen(v) == 5 )) 
{ 
 return 1; 
}

return -1;
}

int main (int argc, char *argv[])
{
FILE *file;
char line[30];

if (( file = fopen("/usr/share/dict/words", "r") ) == NULL) 
{
 fprintf(stderr, "cannot open %s\n", "/usr/share/dict/words");
 exit(1);
} 

while ( fgets(line, sizeof(line), file) )
{

 if ( all_vowels(line) == 1 )
 {
 printf("%s\n", line);
 }
}
fclose(file);
return 0;

}
</code></pre>

blocks|key|4314321|text|我对解析你的文件有点不确定，但是我下面的函数检查一个字符是不是一个元音，并测试下一个元音是否大于当前元音。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4314322|#include+<stdio.h>

//+for+readability+not+advocating+the+
//+usage+of+#define+booleans+etc
#define+TRUE++1
#define+FALSE+0

int+isVowel+(char+c)
{
++++switch+(c)
++++{
++++++++case+'a':+return+TRUE;
++++++++case+'e':+return+TRUE;
++++++++case+'i':+return+TRUE;
++++++++case+'o':+return+TRUE;
++++++++case+'u':+return+TRUE;
++++++++case+'A':+return+TRUE;
++++++++case+'E':+return+TRUE;
++++++++case+'I':+return+TRUE;
++++++++case+'O':+return+TRUE;
++++++++case+'U':+return+TRUE;
++++}

++++return+FALSE;
}

int+hasOrderedVowels+(char+*str)
{++++
++++char+c1,+c2;

++++c1+=+*str;
++++c2+=+*(%2B%2Bstr);

++++//+ignore+words+beginning+in+vowels+other+then+'a'+or+'A'
++++if+(isVowel(c1)+&&+!(c1+==+'a'+%7C%7C+c1+==+'A'))+return+FALSE;

++++do+{
++++++++//+ignore+case+of+`c1`
++++++++if+(c1+>=+'a')
++++++++++++c1+-=+32;

++++++++//+ignore+case+of+`c2`
++++++++if+(c2+>=+'a')
++++++++++++c2+-=+32;

++++++++//+compare+vowels+and+increment
++++++++//+pointers+as+appropriate
++++++++if+(isVowel(c1)+&&+isVowel(c2))
++++++++{
++++++++++++//+if+we+have+found+a+vowel+less+then+or+equal+to+current
++++++++++++//+then+they+are+not+in+order/more+then+one,+if+we+have+found
++++++++++++//+a+'U'+and+there+are+more+vowels+then+this+would+be+a+duplicate
++++++++++++if+(c2+<=+c1+%7C%7C+c1+==+'U')
++++++++++++++++return+FALSE;

++++++++++++c1+=+c2;
++++++++}+
++++++++else+if+(isVowel(c2))++++//+found+first+vowel+so+assign+to+c1
++++++++{
++++++++++++if+(!(c1+==+'a'+%7C%7C+c1+==+'A'))
++++++++++++{
++++++++++++++++return+FALSE;
++++++++++++}
++++++++++++c1+=+c2;
++++++++}
++++++++else+if+(!isVowel(c1))+++
++++++++{
++++++++++++c1+=+*(str+%2B=+2);++++//+skip+over+c2
++++++++}
++++++++c2+=+*(%2B%2Bstr);+
++++}+
++++while+(c2+!=+'\0');

++++return+(c1+==+'U');
}

int+main+()
{
++++char+*str[]+=+{"aeiou",+"facecious",+"chimpanze",+"baboon"};
++++int+i+=+0;

++++for+(;+i<5;+i%2B%2B)
++++{
++++++++printf+("%25s:+%25i\n",+str[i],+hasOrderedVowels(str[i]));
++++}

++++return+0;
}|code-block|syntax|javascript|4314323|demo|offset|length|4314324|entityMap|0|LINK|mutability|MUTABLE|url|http://ideone.com/dPvc3^0|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

little unsure regarding parseing your file yet i function(s) below check that a character is a vowel and tests whether the next vowel is greater then the current vowel. 

<pre><code>#include &lt;stdio.h&gt;

// for readability not advocating the 
// usage of #define booleans etc
#define TRUE 1
#define FALSE 0

int isVowel (char c)
{
 switch (c)
 {
 case 'a': return TRUE;
 case 'e': return TRUE;
 case 'i': return TRUE;
 case 'o': return TRUE;
 case 'u': return TRUE;
 case 'A': return TRUE;
 case 'E': return TRUE;
 case 'I': return TRUE;
 case 'O': return TRUE;
 case 'U': return TRUE;
 }

 return FALSE;
}

int hasOrderedVowels (char *str)
{ 
 char c1, c2;

 c1 = *str;
 c2 = *(++str);

 // ignore words beginning in vowels other then 'a' or 'A'
 if (isVowel(c1) &amp;&amp; !(c1 == 'a' || c1 == 'A')) return FALSE;

 do {
 // ignore case of `c1`
 if (c1 &gt;= 'a')
 c1 -= 32;

 // ignore case of `c2`
 if (c2 &gt;= 'a')
 c2 -= 32;

 // compare vowels and increment
 // pointers as appropriate
 if (isVowel(c1) &amp;&amp; isVowel(c2))
 {
 // if we have found a vowel less then or equal to current
 // then they are not in order/more then one, if we have found
 // a 'U' and there are more vowels then this would be a duplicate
 if (c2 &lt;= c1 || c1 == 'U')
 return FALSE;

 c1 = c2;
 } 
 else if (isVowel(c2)) // found first vowel so assign to c1
 {
 if (!(c1 == 'a' || c1 == 'A'))
 {
 return FALSE;
 }
 c1 = c2;
 }
 else if (!isVowel(c1)) 
 {
 c1 = *(str += 2); // skip over c2
 }
 c2 = *(++str); 
 } 
 while (c2 != '\0');

 return (c1 == 'U');
}

int main ()
{
 char *str[] = {"aeiou", "facecious", "chimpanze", "baboon"};
 int i = 0;

 for (; i&lt;5; i++)
 {
 printf ("%s: %i\n", str[i], hasOrderedVowels(str[i]));
 }

 return 0;
}
</code></pre>

<a href="http://ideone.com/dPvc3" rel="nofollow">demo</a>

So for a lab at uni... Ive been challenged to find all words in the usr/share/dict/linux.words 
file using fopen, fgets etc with every vowel only once, in order.

i.e. facetious

So far I have the following code... but its flawed somewhere...

<pre><code>int all_vowels( char *s )
{ 
 const unsigned char *p = (const unsigned char *)s;

 char *v = malloc(sizeof(char *));
 char *vowel = v;

if(*p == '\0') return -1;

while( *p != '\0' )
{ 
 if( *p == 'a' || *p =='e' || *p =='i'|| *p =='o' || *p =='u' )
 {
 *v = *p;
 v++;
 }
 p++;
 }


 if ( *vowel == 'a' &amp;&amp; (*vowel + 1) == 'e' &amp;&amp; (*vowel + 2) == 'i' &amp;&amp; (*vowel + 3) == 'o' &amp;&amp; (*vowel + 4) == 'u' ) 
 { 
 return 1; 
 }

 return -1;
}

int main (int argc, char *argv[])
{ 
 FILE *file;
 char line[BUFSIZ];

 if (( file = fopen("/usr/share/dict/words", "r") ) == NULL) 
 {
 fprintf(stderr, "cannot open %s\n", "/usr/share/dict/words");
 exit(1);
 } 

 while ( !feof(file) )
 {
 fgets(line, sizeof(line), file);
 if ( all_vowels(line) == 1 )
 {
 printf("%s\n", line);
 }
 }
 fclose(file);
 return 0;

}
</code></pre>

Any tips would be great!!!

Im really confused at the moment...

Find Word with Vowels in order C

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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所以对于大学的实验室来说...我遇到的挑战是在usr/share/dict/linux.words文件中使用fopen、fgets等查找所有单词，每个元音只按顺序查找一次。即搞笑的到目前为止，我有以下代码...但它在某些地方有缺陷..。int all_vowels( char *s ){     const unsig...

问按C顺序查找带有元音的单词
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问按C顺序查找带有元音的单词EN