iOS 转换十六进制值

Question

如何将十六进制值转换为表情符号，我有一个如下所示的字符串

NSString *myVal = @"1F61E";

如何转换此文本以将其显示为表情符号字符？

我找到了 from this link 的值 请告诉我，我真的很纠结这个问题

Updated

NSString *utf8String1 = @"1F61E";
NSString *a = [self convert:utf8String1];
NSLog(@"%@ &&&&&&&&&&&&&&&&&&&&&",a);


-(NSString*)convert:(NSString*)decoded{

    unichar unicodeValue = (unichar) strtol([decoded UTF8String], NULL, 16);
    char buffer[2];
    int len = 1;

    if (unicodeValue > 127) {
        buffer[0] = (unicodeValue >> 8) & (1 << 8) - 1;
        buffer[1] = unicodeValue & (1 << 8) - 1; 
        len = 2;
    } else {
        buffer[0] = unicodeValue;
    }

    return [[NSString alloc] initWithBytes:buffer length:len encoding:NSUTF8StringEncoding];



}

Answer 1

您尝试编码的代码点不适合 16 位。因此需要使用 UTF-32 编码：

NSScanner *scan = [[NSScanner alloc] initWithString:@"1F61E"];
unsigned int val;
[scan scanHexInt:&val];
char cc[4];
cc[3] = (val >> 0) & 0xFF;
cc[2] = (val >> 8) & 0xFF;
cc[1] = (val >> 16) & 0xFF;
cc[0] = (val >> 24) & 0xFF;
NSString *s = [[NSString alloc]
    initWithBytes:cc
           length:4
         encoding:NSUTF32StringEncoding];
NSLog(@"[%@]", s);

Answer 2

您的第一步是将其转换为它的数值：

unichar unicodeValue = (unichar) strtol([input UTF8String], NULL, 16);

那么，按照post的规则：

char buffer[2];
int len = 1;

if (unicodeValue > 127) {
    buffer[0] = (unicodeValue >> 8) & (1 << 8) - 1;
    buffer[1] = unicodeValue & (1 << 8) - 1; 
    len = 2;
} else {
    buffer[0] = unicodeValue;
}

return [[NSString alloc] initWithBytes:buffer length:len encoding:NSUTF8StringEncoding];

您现在有了 UTF-8 格式的字符串！