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Objective-C解码引用的可打印文本

更新时间:2023-02-23 12:58:13

我现在使用C级字符串操作获取了它.它像一种魅力.这样我在iPad2上将测试文件的处理时间从6分钟减少到3秒:

I've got it now using C-Level string manipulation. It works like a charm. This reduced my processing time for a test file from 6 minutes to 3 seconds on an iPad2:

- (char *)replace1:(char const * const)original
{
    char const * const pattern = "=\r\n";

    size_t const patlen = strlen(pattern);
    size_t const orilen = strlen(original);

    size_t patcnt = 0;
    const char * oriptr;
    const char * patloc;

    // find how many times the pattern occurs in the original string
    _FOR(oriptr = original;
         (patloc = strstr(oriptr, pattern));
         oriptr = patloc + patlen)
    {
        patcnt++;
    }

    {
        // allocate memory for the new string
        size_t const retlen = orilen - patcnt * patlen;
        char * const returned = (char *) malloc( sizeof(char) * (retlen + 1) );

        _IF(returned != NULL)
        {
            // copy the original string,
            // replacing all the instances of the pattern
            char * retptr = returned;
            _FOR(oriptr = original;
                 (patloc = strstr(oriptr, pattern));
                 oriptr = patloc + patlen)
            {
                size_t const skplen = patloc - oriptr;
                // copy the section until the occurence of the pattern
                strncpy(retptr, oriptr, skplen);
                retptr += skplen;
            }
            // copy the rest of the string.
            strcpy(retptr, oriptr);
        }
        return returned;
    }
}

- (char *)replace2:(char const * const)original
{
    size_t const replen = 1;
    size_t const patlen = 3;
    size_t const orilen = strlen(original);

    size_t patcnt = 0;
    const char * oriptr;
    const char * patloc;

    // find how many times the pattern occurs in the original string
    _FOR(oriptr = original; (patloc = strstr(oriptr, "=")); oriptr = patloc + patlen)
    {
        patcnt++;
    }

    {
        // allocate memory for the new string
        size_t const retlen = orilen + patcnt * (replen - patlen);
        char * const returned = (char *) malloc( sizeof(char) * (retlen + 1) );

        _IF(returned != NULL)
        {
            // copy the original string,
            // replacing all the instances of the pattern
            char * retptr = returned;
            _FOR(oriptr = original;
                 (patloc = strstr(oriptr, "="));
                 oriptr = patloc + patlen)
            {
                char newRep[3];

                newRep[0] = patloc[1];
                newRep[1] = patloc[2];
                newRep[2] = '\0';

                char _rep[2];
                _rep[0] = (char)(int)strtol(newRep, NULL, 16);
                _rep[1] = '\0';

                size_t const skplen = patloc - oriptr;
                // copy the section until the occurence of the pattern
                strncpy(retptr, oriptr, skplen);
                retptr += skplen;
                // copy the replacement
                strncpy(retptr, _rep, replen);
                retptr += replen;
            }
            // copy the rest of the string.
            strcpy(retptr, oriptr);
        }
        return returned;
    }
}

- (NSString *)decodedQuotedPrintable:(NSString *)string
{
    char *temp = [self replace1:(char *)[string UTF8String]];

    temp = [self replace2:temp];

    return [NSString stringWithCString:temp
                              encoding:NSUTF8StringEncoding];
}