我需要使用UInt8数组,但我不确定如何正确执行此操作...
这是我的代码:
@interface MyClass : NSObject {
__strong id * myArray; //private byte[] myArray; <- Java code
}
@property (nonatomic,readwrite) __strong id * myArray;
@end
这是MyClass中的一个方法:
-(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;
}
这就是我计划调用此方法来填充myBuffer的方法:
UInt8 myBuffer[10000];
[xxx read: myBuffer]; <- how to do this correctly ?????
这是等效的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;
}
这就是我在Java中调用这个方法的方法:
byte[] myBuffer = new byte[10000];
int n = read(myBuffer);
答案 0 :(得分:2)
在Objective-C中,您可以使用NSData
个对象作为字节缓冲区,使用dataUsingEncoding
来获取字符串的字节表示形式:
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