We are running a survey to help us improve the experience for all of our members. If you see the survey appear, please take the time to tell us about your experience if you can.
HAI, I would like to implement TTL interface to a swipe card reader. I want to read track 2 of a card .This contains alphanumeric characters and each character is of 5 bit size which includes one parity bit. I designed a hardware circuit for this with AT89c2051. The data is coming as a stream of bits from the readers magnetic head. So I have store this bit stream to a bit array. I used bdata variable for this purpose, but is not possible to write to the bit position. How I can write these bits to my bit array? What is wrong with code below: I used the pin P1.6 as the input for the data from card reader's magnetic head. Please help me.
sbit dat=P1^6; bdata unsigned char mydata[13]; for (i = 0; i < 13; i++) { for (mask = 0x01; mask != 0x00; mask <<= 1) { mydata[i] & mask = dat; } }
Here the error I got is errorC141:syntax error near'=';
Please give a replay.
Have you tried "myspace" ? :-)
C doesn't has any bit arrays.
Standard C programming can allow you to use a byte array to store any sequence of bits. All you need are the shift operations << and >> together with bit-and & and bit-or |.
To insert five bits at a time, you can keep two variables.
Current start byte to write in. (0 .. sizeof(array)) Current bit position to write in. (0 .. 7)
if bit offset is 0: write only in first byte and increment bit offset by 5.
if bit offset is 1: write only in first byte and increment bit offset by 5.
if bit offset is 2: write only in first byte and increment bit offset by 5.
if bit offset is 3: write only in first byte. then increment byte offset and set bit offset to 0.
if bit offset is 4: write four bits in first byte. then write 1 bit to start of second byte. increment byte offset and set bit offset to 1.
...
if bit offset is 7: write one bit to first byte. then write 4 bits to start of second byte. increment byte offset and set bit offset to 4.
The above can be merged into quite few lines of C code, and is left as an exercise.
Note that writes to the first byte may require a shift left to skip a number of bit positions. Writes to the second byte may require a shift right, to throw away already written bits.
If a 16-bit processor was used, it would normally be efficient to have a 16-bit variable where the data is shifted in, and whenever it contains >= 8 bits of data, then one byte is written. This of course also requires that the transfer is finalized by inserting dummy bits after the transfer, until a full byte is available for storing.
HAI, How can I acess a single bit from a character and store a bit into that location ? can you please post a simple example of that.
Access:
(*Store) & (1<<BitNumber);
Store:
if (BitVal==0) *Store &= ~(1<<BitNumber); else *Store |= (1<<BitNumber);
BitNumber is a value 0 to 7.
Since bit manipulation into normal C data types (char or int) is really standard: Why do you not use Google to search for example code? Are you afraid that you may find too many examples?
read bit 5:
unsigned char c = (mydata[i] & 0x20) >> 5 ;
turning bit 5 on:
mydata[i] |= 0x20 ;
turning bit 5 off:
mydata[i] &= ~0x20 ;
This has two magic constants that needs to be updated if changing to another bit number:
Probably better to only have one constant that depends on which bit position to work with.
unsigned char c = (mydata[i] >> 5) & 1;
I really prefer (1<<5) before 0x20 when working with single bits.