Question

这是其中一项任务，我想知道为什么没有基本的C99功能名称为safeafe（char * in）的地方就是这样做的。这样会很好在一个完美的世界里，对于用UTF-8希伯来语或也许希腊但是假设我被七位ascii卡住了然后我试过了以下内容：

  /* We may even go so far as to ensure that any apostrophe, or hyphen
   * or period may only appear as single entities and not as two or three
   * in a row.  This does not, however, protect us from total non-sense
   * such as D'Archy Mc'Clo.u. d.
   *
   * walk the first_name string and throw away everything that is 
   * not in the range A-Z or a-z, with the exception of space char which
   * we keep. Also a single hyphen is allowed. 
   * 
   * This all seems smart but we are not protected from stupidity such
   * As a name with spaces and dashes or hypens intermixed with letters
   * or from the artist formally known as 'Prince'. 
   */
    char buffer[256];
    j = 0;
    for ( k=0; k<strlen(first_name); k++ ) {

        /* Accept anything in the a - z letters */
        if ( ( first_name[k] >= 'a' ) && ( first_name[k] <= 'z' ) )
            buffer[j++] = first_name[k];

        /* Accept anything in the A - Z letters */
        if ( ( first_name[k] >= 'A' ) && ( first_name[k] <= 'Z' ) )
            buffer[j++] = first_name[k];

        /* reduce double dashes or hyphens to a single hyphen */
        while (( first_name[k] == '-' ) && ( first_name[k+1] == '-' ))
            k++;
        if ( first_name[k] == '-' )  /* do I need this ? */
            buffer[j++] = first_name[k];

        /* reduce double spaces to a single space */
        while (( first_name[k] == ' ' ) && ( first_name[k+1] == ' ' ))
            k++;
        if ( first_name[k] == ' ' )   /* do I also need this ? */
            buffer[j++] = first_name[k];

    }
    /* we may still yet have terminating spaces or hyphens on buffer */
    while ( ( j > 1 ) && (( buffer[j-1] == ' ' ) || ( buffer[j-1] == '-' )) )
        j--;
    buffer[j] = '\0';

    /* Accept this new cleaner First Name */
    strcpy ( first_name, buffer );

只要输入名称缓冲区不超过255，

似乎工作得很好 chars的长度。然而，在第一关，我想知道如何摆脱领先的空间和破折号和连字符以及可能的撇号混合的噪音？

所以问题是......如何使这更好，而且，我是否需要这些线在哪里我问（first_name [k] ==' - '）和空格相同吗？我已经做了走过缓冲区寻找重复，并应该落在连字符上或单个空间。对？

Answer 1

纯粹将其视为如何清理代码的抽象编程问题，您可以使用isalpha()来确保缓冲区仅包含字母，单个连字符和单个空格的字母：

for (k = 0; k < strlen(first_name); k++) {

   if (isalpha (first_name[k])
       buffer [j++] = first_name[k++];

   else if (( first_name[k] == '-' ) && (isalpha (first_name [k+1])))
       buffer [j++] = first_name[k++];

   else if (( first_name[k] == ' ' ) && (isalpha (first_name [k+1])))
       buffer [j++] = first_name[k++];

   else
       k++;
}

这只是一个草案。我实际上没有试过这个，所以没有保证。此外，这不会处理像“约翰 - 保罗”这样的名字在连字符之前和之后用空格写出的情况;你最终会得到一个空格而不是一个连字符。如果你想捕获这样的边缘情况，你可以输入几个额外的“else”子句。

那就是说，作为一个具体的现实世界的解决方案，我同意将名称完全视为输入处理更好。我自己有一个不同寻常的名字，我厌倦了向人们解释是的，这真的是我的名字，不，你不能改变它以适应你可接受的名字的想法。

试图确保字符串缓冲区是一个合理的人名

1 个答案: