I'm trying to get the N leftmost bits from a Core Foundation CFDataRef. New to CF and C, so here's what I've got so far. It's printing out 0, so something's off.

int get_N_leftmost_bits_from_data(int N)
{
    // Create a CFDataRef from a 12-byte buffer
    const unsigned char * _Nullable buffer = malloc(12);
    const CFDataRef data = CFDataCreate(buffer, 12);

    const unsigned char *_Nullable bytesPtr = CFDataGetBytePtr(data);

    // Create a buffer for the underlying bytes
    void *underlyingBytes = malloc(CFDataGetLength(data));

    // Copy the first 4 bytes to underlyingBytes
    memcpy(underlyingBytes, bytesPtr, 4);

    // Convert underlying bytes to an int
    int num = atoi(underlyingBytes);

    // Shift all the needed bits right
    const int numBitsInByte = 8;
    const int shiftedNum = num >> ((4 * numBitsInByte) - N);

    return shiftedNum;
}

Thank you for the help!

CodePudding user response：

Since you're only concerned about the bit in the first four bytes, then you could just copy the bytes over to an integer then perform a bit-shift on that integer.

#include <string.h>
#include <stdint.h>
int main(){
    
    uint8_t bytes[12] = {0};
    for(int n = 0; n < sizeof(bytes) ;   n){
        bytes[n] = n 0xF;
        //printf("x\n",bytes[n]);
    }
    uint32_t firstfour = 0;
    //copy the first four bytes
    memcpy(&firstfour,bytes,sizeof(uint32_t));
    //get the left most bit
    uint32_t bit = firstfour>>31&1;
    
    return 0;
}

You can still perform memcpy in CF

CodePudding user response：

In general, to get n leftmost bits from x, you have to do something like x >> ((sizeof(x) * 8) - n) (I assume that byte consists of 8 bits)

I don't have access to apple's device and don't know its API but few remarks:

• You don't define ret in your's supplied code. (and I don't think it is defined by apple's library)
• I was not able to find CFNumberGetInt32() with my search engine and have no clue what it could do. Please post buildable code with enough context to understand it