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accept user input via scanf() in the form of 4 hexadecimal digits. Your code will interpret this input as a 16-bit unsigned short. For each input item, your code should extract bits 6 through 9 and print out the unsigned numerical value of those bits
I had a sample that says: input is: abcd the result is: 15
i've be working on it but i was converting to decimal from hex and that was a different thing
i just want an explanation about what extract bits 6 to 9 is
0xabcd = 0b1010 1011 1100 1101
bits --------98 7654 3210
bits 6 to 9 11 11
and
0b1111 == 0xf == 15
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What is the meaning of
TIFR0 = 1<<TOV0;
Is it means we are writing 1 to TOV0 bit of TIFR0?
No. That means we are writing 1 to a bit at position TOV0, while writing 0 to all other bits in TIFR0 register.
<< is a bitwise shift left operation, which results equals to all bits of the source number, shifted left in their binary representation.
1 is a number where only the first bit (i.e. the bit at position #0) is set. Therefore 1 << n moves that bit n positions left, and gives you a number where only bit #n is set, but all other bits are cleared.
TOV0 is a macro, which defines the position of bit TOV0.
Be careful when using simple assignment like that, because it will clear all other bits.
If you want to modify the only bit, you may use a bitwise or:
TIFR0 |= 1<<TOV0;
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I'm receiving the following data over a serial port: <0x1b><0x2e><0x15>...
Each value enclosed in '<>' is a single byte.
I require the third byte from the data so i do this:
int Length;
char Data[..];
Length = Data[2];
But the value of Length is 21 and not 15 because the value written in memory is hex.
How do i convert the decimal representation of 15 to decimal 15?
I've tried converting it to various types and so on..
But none of that works for me as i'm writing a driver and performance matters a lot.
I've looked over stackoverflow and other sites but all the given examples are with strings, none are with plain integers.
When i send it to the rest of the algorithm i run into issues as the algorithm expects 15.
Given int x that contains 8 bits that represent a number using natural packed binary-coded decimal, they can be converted to the number with:
int y = x/16*10 + x%16;
x/16 produces the high four bits, and then multiplying by ten scales them to the tens position. x%16 produces the low four bits. They are kept in the ones position and added to the tens.
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Suppose I have the number 0b000 and I need to set the correct bits, so that they will be equal for ex. 5. (0b101).
How do I do that, using bitwise operations algorithm?
Okay, more details then. I'm developing morse code decoder, and to describe a input, I'm using 8 bits: 000 00000, where first three bits are the number of dot/dashes given, and the rest of bits are reserved for the input, where dot is 0, and dash is 1.
For example, the letter A (.-) would be: 010 01000.
The question is, how can I modify the first three bits so that they will show how many dot/dashes were given during the input?
You switch bits on using |. Let's stick with your non-standard notation for binary literals (note that C++14 onwards supports it):
0b000 | 0b100 is 0b100.
0b100 | 0b001 is 0b101.
Note that you can toggle bits using ^ (work through some examples as an exercise).
Finally, you can switch off bits using '&~`.
Solved, if you want to set the first 3 bits, then shift 5 bits to the left (where value is the number you want to set on 3 first bits):
value = value << 5;
And then OR it with the rest of bits:
morseBits = morseBits | value;
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In the function declaration below for uart write on a mcu, can I pass a hex command?
uart_write(const uart_t uart, const uint8_t data);
uart_write(uart_1, 0x56);
yes, you can. However, instead of saying pass a hex command it would be more correct to say pass the value using hex representation.
Anyway
uart_write(uart_1, 0x56);
is the same as
uart_write(uart_1, 86); // 86 == 5 * 16 + 6
An integer can be given in many formats/representations - the compiler just converts the value into a representation suitable for the compiler. Perhaps this could be interesting for you: http://www.cplusplus.com/doc/hex/
If you mean a hexadecimal number, yes you can. To write a hexadecimal number, be sure to precede it with 0x. You can also write octal (base 8) numbers by preceding it with a 0
0x56 Hexadecimal 56
011 Octal 11
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I wonder is there anyway to get the first 2 most significant bits of a binary number version of a decimal number without converting it to actually binary number. For example, I want to get "00" from 5 which is "00101" in binary. Does c supports this binary conversion in an easy way? Does c support shifting and bit masking etc?
Thanks
Yes it does!
So to get the first 2 bits from an 8-bit number we need to mask your current number.
We can do this:
some_number = some_number & 0xC0;
some_number &= 0xC0; // This does the same thing with a different syntax
So what we're doing here is performing a bitwise AND with the value 0xC0, converting 0xC0 to binary gives us 1100 0000. So when we AND this value with some_number we get ONLY the top 2 bits as every other value in the number is 0 and so produces a 0 whenever we AND it with anything else.
For numbers larger than 8-bit all we have to do is increase the length of our mask.
Here's another example say we want the top 2 bits of a 32 bit integer
some_number = some_number & 0xC00000