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iPhone:-[NSConcreteMutableData fastest编码]:无法识别的选择器

nsstring iphone

Isaac Waller · 技术社区 · 17 年前

我正在尝试这样做:

self.somestring = [@"hardcodedstring" stringByAppendingString:someotherstring];

*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '*** -[NSConcreteMutableData fastestEncoding]: unrecognized selector sent to instance 0x1fb930'

我没有在任何地方调用这个方法,但我看到stringByAppendingString:在我的堆栈中调用它:

Stack: (
808221155,
806100816,
808224837,
807957033,
807851552,
812064725,
812064413,
18085, <== -[NSString stringByAppendingString:] + 109 in section LC_SEGMENT.__TEXT.__text of /Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS2.2.1.sdk/System/Library/Frameworks/Foundation.framework/Foundation
817945012,
821160740,
821157652,
821149552,
807852041,
812070519,
807836299,
807834407,
827752032,
816118388,
816157144,
8381,
8244

编辑

@property (nonatomic,retain) NSString *somestring;

@synthesize somestring;

编辑

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingImage:(UIImage *)img editingInfo:(NSDictionary *)editInfo {
    self.somestring = [@"hardcodedstring" stringByAppendingString:[UIImagePNGRepresentation(img) encodeBase64]]; //encodeBase64 encodes it to base64
}

2009-03-20 15:14:21.389 myproject[3467:20b] *** -[NSConcreteMutableData getCharacters:range:]: unrecognized selector sent to instance 0x1fa630
2009-03-20 15:14:21.403 myproject[3467:20b] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '*** -[NSConcreteMutableData getCharacters:range:]: unrecognized selector sent to instance 0x1fa630'
2009-03-20 15:14:21.412 myproject[3467:20b] Stack: (
808221155,
806100816,
808224837,
807957033,
807851552,
807660389,
807733601,
807733297,
807891629,
812155873,
812293801,
18081,
817945012,
821160740,
821157652,
821149552,
807852041,
812070519,
807836299,
807834407,
827752032,
816118388,
816157144,
8381,
8244
)
terminate called after throwing an instance of 'NSException'
(gdb) info symbol 18081
-[InternalChoicesController imagePickerController:didFinishPickingImage:editingInfo:] + 73 in section LC_SEGMENT.__TEXT.__text of /Users/isaacwaller/Documents/myproject/build/Debug-iphoneos/myproject.app/myproject
(gdb) info symbol 812293801
NSLog + 25 in section LC_SEGMENT.__TEXT.__text of /Developer/Platforms/iPhoneOS.platform/Developer/SDKs/iPhoneOS2.2.1.sdk/System/Library/Frameworks/Foundation.framework/Foundation

static const char encodingTable[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";

@implementation NSData (Base64)

- (NSString *)encodeBase64;
{
if ([self length] == 0)
    return @"";

char *characters = malloc((([self length] + 2) / 3) * 4);
if (characters == NULL)
    return nil;
NSUInteger length = 0;

NSUInteger i = 0;
while (i < [self length])
{
    char buffer[3] = {0,0,0};
    short bufferLength = 0;
    while (bufferLength < 3 && i < [self length])
        buffer[bufferLength++] = ((char *)[self bytes])[i++];

    //  Encode the bytes in the buffer to four characters, including padding "=" characters if necessary.
    characters[length++] = encodingTable[(buffer[0] & 0xFC) >> 2];
    characters[length++] = encodingTable[((buffer[0] & 0x03) << 4) | ((buffer[1] & 0xF0) >> 4)];
    if (bufferLength > 1)
        characters[length++] = encodingTable[((buffer[1] & 0x0F) << 2) | ((buffer[2] & 0xC0) >> 6)];
    else characters[length++] = '=';
    if (bufferLength > 2)
        characters[length++] = encodingTable[buffer[2] & 0x3F];
    else characters[length++] = '=';    
}

return [[[NSString alloc] initWithBytesNoCopy:characters length:length encoding:NSASCIIStringEncoding freeWhenDone:YES] autorelease];
}

@end

谢谢,艾萨克·沃勒

2 回复 | 直到 17 年前

Chuck 17 年前

我会检查代码中的拼写错误。该错误非常清楚地表明,在预期使用NSString的地方使用了NSMutableData的实例。似乎encodeBase64正在以某种方式返回 self 否则,原始UIImagePNG表示(img)将被传递给 stringByAppendingString: .

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingImage:(UIImage *)img editingInfo:(NSDictionary *)editInfo {
    NSData *rep = UIImagePNGRepresentation(img);
    NSLog(@"Rep: %@", rep);
    NSString *base64 = [rep encodeBase64];
    NSLog(@"Base 64 is a %@", NSStringFromClass([base64 class]));
    self.somestring = [@"hardcodedstring" stringByAppendingString:base64]; //encodeBase64 encodes it to base64

}

Benjamin Autin Andrés Bonilla 17 年前

~~@"Some String" 不是对象,而是字符串文字。您无法向其发送消息。您需要执行其他操作来加入这些字符串。~~

类似于:

 [NSString stringWithFormat:@"%@%@", @"String 1", @"String 2"];

显然,这是不对的。字符串文字被视为对象。

正如评论中提到的,您也可能遇到以下问题 self.somestring 也。如果你还没有声明属性或合成 somestring 然后通过访问它 self.

somestring = [NSString stringWithFormat:@"%@%@", @"String 1", @"String 2"];

@interface myClass {
  NSString *somestring;
}
@property(nonatomic, retain) NSString *somestring;
@end

@implementation myClass
@synthesize somestring;
@end

self.somestring ,这实际上只是语法糖 [self somestring] .

@property 和 @synthesize

@interface myClass {
  NSString *somestring;
}

-(NSString *)somestring;
-(void)setSomestring:(NSString *)value;
@end

以及:

@implentation myClass
-(NSString *)somestring {
  return somestring;
}
-(void)setSomestring:(NSString *)value {
  somestring = value;
}
@end

somestring 作为一个属性并对其进行合成,那么您就没有这些方法来回答传递的消息。