c - 在 C 中返回 char 数组的长度

Question

我是 C 语言编程的新手，正在尝试编写一个简单的函数来规范化 char 数组。最后我想返回新 char 数组的长度。我来自 Java，所以如果我犯了看似简单的错误，我深表歉意。我有以下代码:

/* The normalize procedure normalizes a character array of size len 
   according to the following rules:
     1) turn all upper case letters into lower case ones
     2) turn any white-space character into a space character and, 
        shrink any n>1 consecutive whitespace characters to exactly 1 whitespace

     When the procedure returns, the character array buf contains the newly 
     normalized string and the return value is the new length of the normalized string.

*/
int
normalize(unsigned char *buf,   /* The character array contains the string to be normalized*/
                    int len     /* the size of the original character array */)
{
    /* use a for loop to cycle through each character and the built in c functions to analyze it */
    int i;

if(isspace(buf[0])){
    buf[0] = "";
}
if(isspace(buf[len-1])){
    buf[len-1] = "";
}

    for(i = 0;i < len;i++){
        if(isupper(buf[i])) {
            buf[i]=tolower(buf[i]);
        }
        if(isspace(buf[i])) {
            buf[i]=" ";
        }
        if(isspace(buf[i]) && isspace(buf[i+1])){
            buf[i]="";
        }
    }

    return strlen(*buf);


}

最后如何返回char数组的长度？另外，我的程序是否正确地完成了我想要的事情？

编辑:我已根据评论对我的程序进行了一些更正。现在正确吗？

/* The normalize procedure normalizes a character array of size len 
   according to the following rules:
     1) turn all upper case letters into lower case ones
     2) turn any white-space character into a space character and, 
        shrink any n>1 consecutive whitespace characters to exactly 1 whitespace

     When the procedure returns, the character array buf contains the newly 
     normalized string and the return value is the new length of the normalized string.

*/
int
normalize(unsigned char *buf,   /* The character array contains the string to be normalized*/
                    int len     /* the size of the original character array */)
{
    /* use a for loop to cycle through each character and the built in c funstions to analyze it */
    int i = 0;
    int j = 0;

    if(isspace(buf[0])){
        //buf[0] = "";
        i++;
    }
    if(isspace(buf[len-1])){
        //buf[len-1] = "";
        i++;
    }
    for(i;i < len;i++){
        if(isupper(buf[i])) {
            buf[j]=tolower(buf[i]);
            j++;
        }
        if(isspace(buf[i])) {
            buf[j]=' ';
            j++;
        }
        if(isspace(buf[i]) && isspace(buf[i+1])){
            //buf[i]="";
            i++;
        }
    }

    return strlen(buf);


}

Answer 1

执行此类操作的规范方法是使用两个索引，一个用于读取，一个用于写入。像这样:

int normalizeString(char* buf, int len) {
    int readPosition, writePosition;
    bool hadWhitespace = false;
    for(readPosition = writePosition = 0; readPosition < len; readPosition++) {
        if(isspace(buf[readPosition]) {
            if(!hadWhitespace) buf[writePosition++] = ' ';
            hadWhitespace = true;
        } else if(...) {
            ...
        }
    }
    return writePosition;
}

警告:这仅根据给定的长度处理字符串。虽然使用缓冲区 + 长度具有能够处理任何数据的优势，但这不是 C 字符串的工作方式。 C 字符串在其末尾以空字节终止，您的工作是确保空字节位于正确的位置。您提供的代码不处理空字节，我上面提供的缓冲区 + 长度版本也不处理。这种规范化函数的正确 C 实现如下所示:

int normalizeString(char* string) {    //No length is passed, it is implicit in the null byte.
    char* in = string, *out = string;
    bool hadWhitespace = false;
    for(; *in; in++) {    //loop until the zero byte is encountered
        if(isspace(*in) {
            if(!hadWhitespace) *out++ = ' ';
            hadWhitespace = true;
        } else if(...) {
            ...
        }
    }
    *out = 0;    //add a new zero byte
    return out - string;    //use pointer arithmetic to retrieve the new length
}

在这段代码中，我用指针替换了索引，只是因为这样做很方便。这只是风格偏好的问题，我可以用明确的索引写同样的东西。 (而且我的风格偏好不是指针迭代，而是简洁的代码。)