Question

我正在尝试这样做：

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

但我一直在接受：

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

我不会在任何地方调用此方法，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

）我如何解决这个问题，以便它附加字符串，就像它应该的那样。谢谢，
艾萨克
修改：要访问somestring，我会这样做：

@property (nonatomic,retain) NSString *somestring;

在我的ViewController.h中：

@synthesize somestring;

在我的ViewController.m中编辑：someotherstring来自这里：

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

更多编辑：我把它分解了：

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

所以我认为编码Base64数据有问题吗？我正在使用的代码是：

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，Isaac Waller

Answer 1

我会检查拼写错误的代码。该错误非常清楚地表明在期望NSString的地方使用NSMutableData的实例。似乎encodeBase64以某种方式返回self，否则原始UIImagePNGRepresentation（img）将传递给stringByAppendingString:。

如果不明显，只需将其分解并检查每一步。（我在这个例子中使用的是NSLog，但当然，使用调试器也可以正常工作。）

- (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

}

我最好的猜测是你不小心传入原始的NSData而不是NSString - 但如果不这样做，上面的代码应该正常工作或准确显示事情发生的地方。

Answer 2

~~@"Some String"不是对象，它是一个字符串文字。您无法向其发送消息。您将需要执行其他操作来加入这些字符串。~~

类似的东西：

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

<击> 显然，这是不正确的。字符串文字被视为对象。

正如评论中所述，您也可能遇到self.somestring的问题。如果您尚未声明属性或已合成somestring，则通过self.访问该属性不正确。你应该使用

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

如果你做完了：

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

和

@implementation myClass @synthesize somestring; @end

然后你可以使用self.somestring访问变量，这实际上只是[self somestring]的语法糖。

重要的是要注意@property和@synthesize本身实际上是语法糖位。他们最终会做类似的事情。

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

和

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

因此，如果您尚未将somestring声明为属性并将其合成，那么您没有这些方法来回答传递的消息。

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

2 个答案: