传递（UInt8数组的指针）和访问数组

Question

目前正在做关于UInt8数组的相关应用，遇到问题。

我的代码：

@interface MyClass : NSObject {
    __strong id * myArray; //private byte[] myArray;  <- Java code
}
@property   (nonatomic,readwrite) __strong id * myArray;
@end

在MyClass中的方法：

-(int) getArray: (__strong id *) bufferTmp {

    NSString* aString = @"theString";
    int bytes  = aString.length;

    //now I need to fill the passed in array with the chars of the String
    for (int i = 0; i < bytes; i++) {
            char c = [aString characterAtIndex:i];
            ??? bufferTmp[i] = (UInt8)c;   <----- what to write here?
        }
return bytes;
}

下面是我准备调用方法充填myBuffer的代码：

UInt8 myBuffer[10000];
[xxx read: myBuffer];       <-      不知道这段正确么?

这是相同的java代码：

public int getArray(byte[] bufferTmp) {
    String theString = "theString";
    for (int i = 0; i < bytes; i++) {
        char c = theString.charAt(i);
        bufferTmp[i] = (byte) c;
        }
     return bytes;

 }

在java中调用方法的代码：

byte[] myBuffer = new byte[10000]; 
int n = read(myBuffer);

Answer 1

在Objective-C中可以使用NSDate对象当做byte buffer, 然后dataUsingEncoding获取字节表示的字符串：

NSString *aString = @"theString";
NSData *myBuffer = [aString dataUsingEncoding:NSUTF8StringEncoding];

const char *bytes = [myBuffer bytes]; // pointer to the bytes in the buffer
NSUInteger count = [myBuffer length]; // number of bytes in the buffer