我有一个带有十六进制字符串的NSString,如#34; 68656C6C6F"这意味着"你好"。
现在我想将十六进制字符串转换为另一个显示" hello"的NSString对象。怎么做?
答案 0 :(得分:20)
我确信有更好,更聪明的方法可以做到这一点,但这个解决方案确实有效。
NSString * str = @"68656C6C6F";
NSMutableString * newString = [[[NSMutableString alloc] init] autorelease];
int i = 0;
while (i < [str length])
{
NSString * hexChar = [str substringWithRange: NSMakeRange(i, 2)];
int value = 0;
sscanf([hexChar cStringUsingEncoding:NSASCIIStringEncoding], "%x", &value);
[newString appendFormat:@"%c", (char)value];
i+=2;
}
答案 1 :(得分:3)
这应该这样做:
- (NSString *)stringFromHexString:(NSString *)hexString {
// The hex codes should all be two characters.
if (([hexString length] % 2) != 0)
return nil;
NSMutableString *string = [NSMutableString string];
for (NSInteger i = 0; i < [hexString length]; i += 2) {
NSString *hex = [hexString substringWithRange:NSMakeRange(i, 2)];
NSInteger decimalValue = 0;
sscanf([hex UTF8String], "%x", &decimalValue);
[string appendFormat:@"%c", decimalValue];
}
return string;
}
答案 2 :(得分:0)
extension String {
func hexToString()-> String {
var newString = ""
var i = 0
while i < self.count {
let hexChar = (self as NSString).substring(with: NSRange(location: i, length: 2))
if let byte = Int8(hexChar, radix: 16) {
if (byte != 0) {
newString += String(format: "%c", byte)
}
}
i += 2
}
return newString
}
}
答案 3 :(得分:-1)
+(NSString*)intToHexString:(NSInteger)value
{
return [[NSString alloc] initWithFormat:@"%lX", value];
}
答案 4 :(得分:-1)
我认为建议initWithFormat的人是最好的答案,因为它的目标是C而不是ObjC,C的混合......(虽然示例代码有点简洁)..我做了以下
unsigned int resInit = 0x1013;
if (0 != resInit)
{
NSString *s = [[NSString alloc] initWithFormat:@"Error code 0x%lX", resInit];
UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Initialised failed"
message:s
delegate:nil
cancelButtonTitle:@"OK"
otherButtonTitles:nil];
[alert show];
[alert release];
[s release];
}