Question

这是我的另一个愚蠢的问题！

NSString *hex1 = @"50be4f3de4";
NSString *hex2 = @"30bf69a299";

/* some stuff like result = hex1^hex2;    */

NSString *result = @"6001269f7d";

我有一个十六进制值作为字符串，存储在两个diff中。变量。我需要Xor他们，结果应该在另一个字符串变量？

我通过转换字符串尝试了它们 - ＆gt; NSData - ＆gt; bytes array - ＆gt; xor'ing他们......但我没有成功..... 提前谢谢你......

Answer 1

您必须先将每个字符转换为Base16（十六进制）格式。然后您应该继续对这些字符进行异或。您可以使用strtol（）函数来实现此目的。

NSString *hex1 = @"50be4f3de4";
NSString *hex2 = @"30bf69a299";

NSMutableArray *hexArray1 = [self splitStringIntoChars:hex1]; 
NSMutableArray *hexArray2 = [self splitStringIntoChars:hex2]; 

NSMutableString *str = [NSMutableString new];
for (int i=0; i<[hexArray1 count]; i++ )
{
    /*Convert to base 16*/
    int a=(unsigned char)strtol([[hexArray1 objectAtIndex:i] UTF8String], NULL, 16);
    int b=(unsigned char)strtol([[hexArray2 objectAtIndex:i] UTF8String], NULL, 16);

    char encrypted = a ^ b;
    NSLog(@"%x",encrypted);
    [str appendFormat:@"%x",encrypted];        
}
NSLog(@"%@",str);

我用来分割字符串

字符的实用方法

-(NSMutableArray*)splitStringIntoChars:(NSString*)argStr{
    NSMutableArray *characters = [[NSMutableArray alloc] 
                              initWithCapacity:[argStr length]]; 
    for (int i=0; i < [argStr length]; i++) 
    { 
        NSString *ichar = [NSString stringWithFormat:@"%c", [argStr characterAtIndex:i ]]; 
        [characters addObject:ichar]; 
    } 
    return characters;
}

希望它有所帮助!!

XOR'ing存储为NSString的两个十六进制值？

1 个答案: