将二进制的“字符串”转换为文本的NSString

Question

我能够将（ASCII）文本的NSString转换为二进制数字的NSString，但我遇到了相反的麻烦。例如：“Hi”变成“01101000 01101001”。

I need: "01101000 01101001" to become "Hi". 

我在寻找最直接的方式来实现这一点。请注意每8位二进制数字之间的空格。

Answer 1

考虑到格式总是这样，此代码应工作：

NSString * 
BinaryToAsciiString (NSString *string) 
{ 
    NSMutableString *result = [NSMutableString string]; 
    const char *b_str = [string cStringUsingEncoding:NSASCIIStringEncoding]; 
    char c; 
    int i = 0; /* index, used for iterating on the string */ 
    int p = 7; /* power index, iterating over a byte, 2^p */ 
    int d = 0; /* the result character */ 
    while ((c = b_str[i])) { /* get a char */ 
     if (c == ' ') { /* if it's a space, save the char + reset indexes */ 
      [result appendFormat:@"%c", d]; 
      p = 7; d = 0; 
     } else { /* else add its value to d and decrement 
        * p for the next iteration */ 
      if (c == '1') d += pow(2, p); 
      --p; 
     } 
     ++i; 
    } [result appendFormat:@"%c", d]; /* this saves the last byte */ 

    return [NSString stringWithString:result]; 
} 

告诉我，如果它的某些部分是不清楚。

Answer 2

这是怎么回事？

NSString* stringFromBinString(NSString* binString) { 
    NSArray *tokens = [binString componentsSeparatedByString:@" "]; 
    char *chars = malloc(sizeof(char) * ([tokens count] + 1)); 

    for (int i = 0; i < [tokens count]; i++) { 
     const char *token_c = [[tokens objectAtIndex:i] cStringUsingEncoding:NSUTF8StringEncoding]; 
     char val = (char)strtol(token_c, NULL, 2); 
     chars[i] = val; 
    } 
    chars[[tokens count]] = 0; 
    NSString *result = [NSString stringWithCString:chars 
              encoding:NSUTF8StringEncoding]; 

    free(chars); 
    return result; 
} 

（发布为社区维基 - 我的老斯库尔ç技能是死的生疏 - 随时清理它:-)）