I'm having a hard time understanding the logic behind successfully setting a bit in a 32 bit register. Here is the pseudo-code for the function:
Read the master register,
If the 29th bit CREG_CLK_CTRL_I2C0 is not set, set it
uint32_t creg;
//read the CREG Master register
creg = READ_ARC_REG((volatile uint32_t)AR_IO_CREG_MST0_CTRL);
if((creg & (1 << CREG_CLK_CTRL_I2C0)) == 0){
creg |= ( 1 << CREG_CLK_CTRL_I2C0);
WRITE_ARC_REG(creg, (volatile uint32_t)(AR_IO_CREG_MST0_CTRL));
}
If the CREG master register is initially empty, the logic doesn't work as intended. However, if I fill it with all zeros and a 1 in the 31st bit (1000...0) the logic does work. I'm not sure if my test condition is incorrect or if it could be something else.
Can anyone help?
Personally, I would use the data type given: uint32_t. That will guarantee alignment regardless of context. That is (factored for clarity, and assuming shift doesn't yield different size-type):
uint32_t mask = ((uint32_t)1) << CREG_CLK_CTRL_I2C0;
if((creg & mask) == 0){
creg |= mask;
WRITE_ARC_REG(creg, (volatile uint32_t)(AR_IO_CREG_MST0_CTRL));
}
Here are a few ideas towards debugging your issue:
Step 1: Make sure you actually understand how that register works. Remember, microcontroller registers may behave differently from memory. The register may ignore your attempt to write a 1 to a bit until after some other condition is met. Perhaps that's why it works if you write a 1 to bit 31 first. What does bit 31 do?
Step 2: I poked around online and found that the same header that defines READ_ARC_REG() and WRITE_ARC_REG() may also include the definition for SET_ARC_BIT(). See if you can find and use that.
Step 3: Make sure what you're trying to write makes sense. Step through the function in your debugger and/or add some form of printout to display the value you're attempting to write to the register. Then read the register after doing so and repeat that process. See if you tried to write the correct value, then see whether that write actually took. If your code tried to write your desired value to the register but the subsequent read showed that your write didn't change the bit then go back to Step 1 above.
Just a guess (I don't do much embedded programming), but according to C standard the number literal 1 has type int. In embedded programming int can be 16 bits, in which case your left shift will have undefined behaviour, since right operand is too large.
So try using 1L to make this of type long.
Also, as mentioned by Olaf in comments, do not use (volatile uint32_t) cast unless you are sure that you need it. Some googling suggests that this is about Arduino 101, and source code I found does use both READ_ARC_REG and WRITE_ARC_REG without any cast.
creg = READ_ARC_REG(AR_IO_CREG_MST0_CTRL);
if((creg & (1L << CREG_CLK_CTRL_I2C0)) == 0){
creg |= (1L << CREG_CLK_CTRL_I2C0);
WRITE_ARC_REG(creg, AR_IO_CREG_MST0_CTRL);
}
Related
Take macro:
GPIOxMODE(gpio,mode,port) ( GPIO##gpio->MODER = ((GPIO##gpio->MODER & ~((uint32_t)GPIO2BITMASK << (port*2))) | (mode << (port * 2))) )
Assuming that the reset value of the register is 0xFFFF.FFFF, I want to set a 2 bit width to an arbitrary value. This was written for an STM32
MCU that has 15 pins per port. GPIO2BITMASK is defined as 0x3. Is there a better way for clearing and setting a random 2 bits in anywhere in the
32-bit wide register.
Valid range for port 0 - 15
Valid range for mode 0 - 3
The method I came up with is to bit shift the mask, invert it, logically AND it with the existing register value, logically OR the result with a bit shifted new value.
I am looking to combine the mask and new value to reduce the number of logical operations bit shift operations. The goal is also keep the process generic enough so that I can use for bit operations of 1,2,3 or 4 bit widths.
Is there a better way?
In the long and sort of it, is there a better way is really an opened question. I am looking specifically for a method that will reduce the number of logical operations and bit shift operations, while being a simple one lined statement.
The answer is NO.
You MUST do reset/set to ensure that the bit field you are writing to has the desired value.
The answers received can be better (in a matter of opinion/preference/philosophy/practice) in that they aren't necessary a macros and have have parameter checking. Also pit falls of this style have been pointed out in both the comments and responses.
This kind of macros should be avoided as a plaque for many reasons:
They are not debuggable
They are hard to find error prone
and many other reasons
The same result you can archive using inline functions. The resulting code will be the same effective
static inline __attribute__((always_inline)) void GPIOMODE(GPIO_TypeDef *gpio, unsigned mode, unsigned pin)
{
gpio -> MODER &= ~(GPIO_MODER_MODE0_Msk << (pin * 2));
gpio -> MODER |= mode << (pin * 2);
}
but if you love macros
#define GPIOxMODE(gpio,mode,port) {volatile uint32_t *mdr = &GPIO##gpio->MODER; *mdr &= ~(GPIO_MODER_MODE0_Msk << (port*2)); *mdr |= mode << (port * 2);}
I am looking to combine the mask and new value to reduce the number of
logical operations bit shift operations.
you cant. You need to reset and then set the bits.
The method I came up with is to bit shift the mask, invert it,
logically AND it with the existing register value, logically OR the
result with a bit shifted new value.
That or an equivalent is the way to do it.
I am looking to combine the mask and new value to reduce the number of
logical operations bit shift operations. The goal is also keep the
process generic enough so that I can use for bit operations of 1,2,3
or 4 bit widths.
Is there a better way?
You must accomplish two basic objectives:
ensure that the bits that should be off in the affected range are in fact off, and
ensure that the bits that should be on in the affected range are in fact on.
In the general case, those require two separate operations: a bitwise AND to force bits off, and a bitwise OR (or XOR, if the bits are first cleared) to turn the wanted bits on. There may be ways to shortcut for specific cases of original and target values, but if you want something general-purpose, as you say, then your options are limited.
Personally, though, I think I would be inclined to build it from multiple pieces, separating the GPIO selection from the actual computation. At minimum, you can separate out a generic macro for setting a range of bits:
#define SETBITS32(x,bits,offset,mask) ((((uint32_t)(x)) & ~(((uint32_t)(mask)) << (offset))) | (((uint32_t)(bits)) << (offset)))
#define GPIOxMODE(gpio,mode,port) (GPIO##gpio->MODER = SETBITS32(GPIO##gpio->MODER, mode, port * 2, GPIO2BITMASK)
But do note that there appears to be no good way to avoid such a macro evaluating some of its arguments more than once. It might therefore be safer to write SETBITS32 as a function instead. The compiler will probably inline such a function in any case, but you can maximize the likelihood of that by declaring it static and inline:
static inline uint32_t SETBITS32(uint32_t x, uint32_t bits, unsigned offset, uint32_t mask) {
return x & ~(mask << offset) | (bits << offset);
}
That's easier to read, too, though it, like the macro, does assume that bits has no set bits outside the mask region.
Of course there are other, similar formulations. For instance, if you do not need to support discontinuous bit ranges, you might specify a bit count instead of a bit mask. This alternative does that, protects against the user providing bits outside the specified range, and also has some parameter validation:
static inline uint32_t set_bitrange_32(uint32_t x, uint32_t bits, unsigned width,
unsigned offset) {
if (width + offset > 32) {
// error: invalid parameters
return x;
} else if (width == 0) {
return x;
}
uint32_t mask = ~(uint32_t)0 >> (32 - width);
return x & ~(mask << offset) | ((bits & mask) << offset);
}
I'm pretty new to bitwise and all the fun jazz and so don't quite understand everything about it. I have two questions.
A) A flags and bitshift question
I recently ran across something similar to below
if (flags & (1 << 3)) {
function_A();
}
I can see is is an AND operator and a left bit shift, however I am unsure what the flag does and its purpose (to my understanding its a collection of booleans to save space), when I usually come across left shifts, it is something such as 10100101 << 3, which would be 00101000 (I believe), but that does not seem to be the case here. So what exactly are the conditions under which the above function would be called?
B) Also a flags question (related to the first due to the nature of it).
TCPs contain packets which consist of 1 bit flags in byte 13.There is a bit of byte 13 (bit 1 i believe) which is the SYN flag to request a connection. To "request a connection" how exactly would you call that bit assuming you can access it assuming its stored in some sort of array and is accessed VIA packetNO[13]. Would it be similar to below?
if (packetNO[13] & (1 << 2)) {
}
the above checking if a connection has been requested, by shifting a true bit to position 2 (bit 1?)
Please explain these concepts to me and provide examples to assist if possible, I am unsure if I am correct or not.
The and operator is such its output is at one only if both operands are at 1.
Hence
if(f & 1) { ... }
tests is the least significant bit of f is set.
If you want to test if another bit is set, there are two ways to do that.
use the bitwise shift operator << that will shift its operand by a given amount. For instance, to test if the third bit (or bit #2 counting from lsb) is set, you an use 1<<2. This will result to a number equal to 000..00100 and by anding, this will check if the corresponding bit is set.
if(f & (0x1<<2)) { ... }
Alternatively, you can use hexadecimal numbers do describe the bit pattern that you want to test.
The same test can done by using 0x4 as the binary code of 4 is 000..0100
if(f & 0x4) { ... }
It is up to you to determine which one is more readable.
So, the first test in your question checks if fourth bit of flag (bit #3) is set and the second one tests if bit #1 of packect[13] is set.
how is:
GIMSK |= (1 << PCIE);
PCMSK |= (1 << PCINT4);
equal to (I can use the above or the below in my setup of my program, both work and activate pin 4), the GIMSK and the PCMSK are for some reason equal to each other, I am trying to learn why.
GIMSK = 0b00100000;
PCMSK = 0b00010000;
first:
https://thewanderingengineer.com/2014/08/11/pin-change-interrupts-on-attiny85/
second:
https://embeddedthoughts.com/2016/06/06/attiny85-introduction-to-pin-change-and-timer-interrupts/
data:
http://ww1.microchip.com/downloads/en/DeviceDoc/Atmel-2586-AVR-8-bit-Microcontroller-ATtiny25-ATtiny45-ATtiny85_Datasheet.pdf
The sheet says PCIE is 0b00100000 in the bit mask, so somehow |= (1 << PCIE) equals that? I don't get it if PCIE is supposed to be that, doing a shift would change that value..
how and why would you use that instead of binary? I would guess it changes it but obviously, somehow it doesn't. I've asked this on several different places nobody has an answer so I came here. Hopefully someone can explain.
I'm new to C, I just learned bitwise operations today to try to figure out what is going on here, my code does work with either or but I want to know why! Thanks.
It is equal because all other bits of those registers were 0 before the OR operation
1u << x shifts one by x positions left. As a result you have the number with all bits except x cleared
Short Version
Break apart the compound expression from x |= y to x = x | y
Load integer '1' which is just binary `[0000 0001] and shift it to the spot we want.
Variable is defined in a header, or elsewhere which says where that bit is for compatibility. New card? Just get a new definition file- this is done automatically- usually). PCIE and PCINT4 are defined somewhere, look for it if needed, but this is supposed to handle those details for you.
Yet- we know is PCINT4 is 4 and PCIE is 5 respectively because of the second set- where we see a shifted '1'- 5 spaces for PCIE and 4 spaces for PCINT4. That's why they seem equivalent. Because they are literally equivalent- once you evaluate the expression you know that 0b means that what follows is binary (which might have been obvious). But they are NOT equivalent. Not exactly- see below--
4-So if you "OR" a register, it overrides whatever is there if TRUE and forces it to be what True and leaves everything else alone. We shift that 1 to show the bit we want, then we choose the operant that will have the effect we want. Look at the table for others.**
But we get GIMSK = GIMSK OR 0b00100000 and PCMSK = PCMSK OR 0b00010000;
Which is similar, but not exactly the same thing.
The devil is in the details, see below for
Detailed Explanation.
AKA someone better read this, took me forever.
GIMSK |= (1 << PCIE);
PCMSK |= (1 << PCINT4);
equal to (I can use the above or the below in my setup of my program, both work and activate pin 4), the
GIMSK and the PCMSK are for some reason equal to each other, I am
trying to learn why.
GIMSK = 0b00100000;
PCMSK = 0b00010000;
Let's take it lexicographically by the token to start, so we are all talking about the same thing.
GIMSK |= (1 << PCIE);
Starting at the first bit. Pick it apart-
GIMSK- variable or 'id'
|= - operator and assignment combo bitwise OR and =
1 - the integer 1
<< - shift operation
PCIE - Variable and ID
Of course, this makes it harder to explain. What is |=? I'm certain that led to confusion on this question for some. The better-known one is +=. So if I have a variable x, and I always add to itself, really any time you are counting, etc. the variable is on both sides of the equation. Like this:
x = x + 1 ;this is so common though, that in C, it was shortened to +=
x += 1 ; now its written like this. It takes some time
so programmers are lazy, and if something 'cool' pops up in one language, it usually spreads to the others, so most languages allow this now. It does make a difference if you write += or =+, at least in Java. Won't hit that though.
y = y * 2; it works for other types of operands
y *= 2; Now we take y, multiply by 2 and assign back to y.
Now, let's look at 'C' style bitwise operators- most languages have adopted similar notation though there are exceptions.
Usually in C you use two symbols to compare '&&" or '||' or '=='
Well, that's because the single operator compares bits. This got one symbol because it's much more natural on a computer and much more common. Not to us anymore, we are abstracted away from it by layers of software.
So we have: ** Good Source for more info
Bitwise AND (&)
Bitwise OR (|)
Bitwise XOR (^)
Bitwise NOT (~)
And we can also make this compond (Click for more info)
Basically, they compare some variable on the right with the left and assign it back to the right. Like this x = x* y => x *= y.
Likewise we have x &= y, x |= y and, x ^= y
So for the above- let's unwrap it first- write it out longhand to make it easier to understand-
GIMSK |= (1 << PCIE)
GIMSK = GIMSK | (1 << PCIE) #OK! much easier to understand if your new.
#NOW we can lexigraphically analyze this
VarA {assignment} VarA OR ( 1 {Operator} VarB )
#Ignore the assignment side, for now, practice order of Operations
#Start with Parenthetical Exp.
1 {Operator} VarB
#It turns out this is defined.
#OP Didnt know but computer does. = 5 in this case.
#so 1, shift left 5. To bitwise shift, need bits
1 => 0b00000001 << 5 = 0b00100000
# shift left is really multiplied by 2 in base 10, divide by 2 in shift right. Beware Right Shift, esp in float.
So now we have: GIMSK = GIMSK OR 0b00100000
GIMSK |= (1 << PCIE); PCMSK |= (1 << PCINT4);
GIMSK = 0b00100000; PCMSK = 0b00010000;
Which is just what you already said. More or less. The 2nd operations are not equivalent though as I mentioned above in the short answer. Thats covered at the end.
This is Assembly format, GIMSK is an 8-bit register. We created a bitmask, by moving a 1 to the register we want to effect, and putting a 0 in the bits we want to leave alone. The |= means we will compare the two the save it back to the same register. That's it. OR 1 will always turn it on. Which is what we want.
Think about what we want to do to start. We want to set a boolean value to HIGH or TRUE, or 1, however, you put it. We say "Lets set the register bit that we specify if your value (0/1) OR my value (1) is 1." Well, we know our value is 1 because that's what we put. So when you bit-wise OR (is that a verb?), you are writing a value on the basis of one of 2 values being a 1. It either writes a one or leaves a one, unless you send 0 and it sends 0 it stays off.
It says "I think this should be on. If any other process thinks this should be on, leave it even if I don't need it so (it sends a 0)" It's worth thinking through on paper, and thinking through each of the operands. Make a colored table, that's how I got to understand them. Not of the operands will flip the values whatever it is. AND checks the value for you, it leaves the register the same- Operating on each reg will have this effect. I used to have a cheat sheet, I would have loved to include, but I lost it- but it summarized for my dumb brain the behavior of each operand.
REMEMBER- we can not operate on a single bit. This is a critical bit you need to understand. You cant change just one bit. If I have 0010 0010 and I want to say, hey computer, change byte 6. You cant! You have to load the whole word or byte into a register or at least half (16bits in MIPS, 8 in ATMEL 16bit controllers**), and operate on the whole thing. You can't operate directly from memory (Ram, SSD, L2 Cache- way too far away). There's no such thing as popping a single bit into a register to change it, though there are tricks to make new bytes (8bit) in the shape you want. Want just the 6th bit, well {0100 000} -with AND, will get it for you. Then you can shift right, or divide by 2^6, etc. We will get back to this. First- the actions of the Comparators if you care to learn more:
*this chip is 8 bit. Doubt they have half read. Bit
Logical Operators v. Registers
Register(b) Me OR NOR XOR AND NAND N XNOR
1 0 1 0 1 0 1 0 0
0 0 0 1 0 0 1 1 1
1 1 1 0 0 1 0 0 1
0 1 1 0 1 0 1 1 0
So the above looks at the single bit we want to effect. The left 2 columns are all possible scenarios (just 4), where we show the bit of the register in that byte. I keep saying byte. A register is, as I mentioned 16 or 32 bits usually, so I really mean Word. I have just organized this example around a hypothetical 8 bit machine. edit- this is an 8 bit chip, 32 registers. one of which is this one
Now! What do we want to do? We Want to change 1 byte, which represents a boolean value, but we don't want to mess with the rest! If you OR across all 8 bytes- do this on paper- it leaves the values in there already alone. Perfect! That's what we want.
Whatever they are set to, they stay, if it's 1, OR leaves it a 1, if it's 0, it stays.
Ah, I should mention why.
Its because you start with 0000 0001 (1) so everything is 0, except the 1st bit. Why did we start with 0000 0001? because you told it to.
See here syntax Arduino Doc Bitwise Ops
So, without reviewing Binary, 1 in binary is 0000 0001 It should be noted, that in a computer, it can't tell that 11001100 isn't 11,001,100 (eleven million), or if its 204 (binary) or even 285,217,024 (if it was HEX), or 2,359,872 (in Octal).
The compiler and computer 'know' we always think base 10, but the computer never does, just base 2 or compressed for easy human reading, into octal (2^2,) or hex (2^4) eg, each 'character' is 2 bits or 4 bits. 0x0A is 0b1010. And right there is where I am getting at. We indicate the values are not base 10, with a prefix. 0b***** is binary. 0x**** is hex. And I can never remember Octal- no one uses it anyway.
So!
see here if needed: Another practical book I wrote on Bitwise Ops, that covers basic Binary a little.
Then you shift that bit by the PIN NUMBER I don't know the right term, and this is certainly not it, but you say the register you want to effect is the 5th register. Ok.
#take a 1,
0000 0001 = $temp
#shift it 5 spots "<<" , where 5 is the PCIE 'bit' value spot number.
1<<5 = 32
#binary equals 32.
You could replace either GIMSK value with 32 and it would be fine, again equivalent, or 0x020
# 0010 0000
# Then OR this with whats in the register now:
1010 1010 (made up number, a mix of ones and 0s)
0010 0000 (Our Value)
OR=>
1010 1010 Result.
Note how we left the other bits alone, and only changed what we wanted! Effective bit mask!
Now, why does it say PCIE, and whatever the other one is. Its because somewhere, when you compile, there is a file that assigns values to those variables. This allows the code to be compatible across several different chip designs. The ATMEGA and the ATTINY do not have the same interrupt pin. Though it likely goes to the same internal register.
#Take it bit by bit, no Pun intended
GIMSK |= (1 << PCIE);
PCMSK |= (1 << PCINT4);
GIMSK = 0b00100000;
PCMSK = 0b00010000;
Again, starting at the first bit from above.
GIMSK- some variable
|= - bitwise OR
1 - the integer 1
`<`< - shift operation
PCIE - another var
So all you are doing is taking a base 10 integer- 1, which we know equals 0b0000 0001, then we are pushing that 1 (now in binary to the spot indicated by PCIE or PCINT4. So the latter 2 are just simply variables that hold the bit number, so if it changes, the code doesn't break.
From the latter 2 lines, we infer that PCIE is 5 and PCINT4 is 4. GIMSK is now equal to 32 and the other 16. Shifting << and >> has the effect of multiplying or diving by 2. Although, shifting down is risky for reasons I won't get into, but if you need to multiply a number by 2, for a computer, it's much faster to shift left by 1 bit than it is to go through the multiplayer.
We talked about the OR already. It sets a 1, if there's not one, otherwise it leaves the other bits alone because they are 0.
Equivalent or Not??
GIMSK |= (1 << PCIE);
GIMSK = 0b00100000;
GIMSK = GIMSK OR 0b00100000
PCMSK |= (1 << PCINT4);
PCMSK = 0b00010000;
PCMSK = PCMSK OR 0b00010000
So evaluating the 1st expression in each set gets the 3rd equation. But notice they look a little different. They are not equivalent statements, though, as you say they may work. It depends on what those other bits are.
PCMSK = 0b00010000; #This sets the PCMSK register to be exactly
=> PCMSK = `0|0|0|1|0|0|0|0
#While
PCMSK = PCMSK OR 0b00010000; # yields PCMSK = `?|?|?|1|?|?|?|?`
#Obviously,
GIMSK = 0b00001000; # This sets the GIMSK register to be exactly
=> GIMSK = `0|0|0|0|1|0|0|0`
While`GIMSK = GIMSK OR 0b00001000; # yields
GIMSK = ` ?|?|?|?|1|?|?|? `
while the OR statement leaves the other bits alone, if they were set by something else, and just changes the 5 (or 4th bit) as the case my be. The OR statement is probably the better statement. If you found the latter statement suggested in a reputable place though, it's probably fine.
Conclusion
So that's it. It's much easier than you thought probably now that the different bits make sense. I wrote this though with the hope it'll give some lasting insight rather than just a quick answer. Although in truth- it was complicated. There are a LOT of computer science concepts buried in those 2 statements, that if you're not in the know, might as well be hieroglyphics.
All this makes much more sense if you dive into how a computer works.
Check out Chapter 2 and 3 of Computer Organization and Design (5th ed) Patterson and Hennesy. It's the standard. If this is for fun, you can skim it. But the computer has Registered, of a defined width- 8, 16, 32. and 64 or even 128 (rarely e.g. x86 AVX- Intel x86). but usually 32 bit. These are the bits of data in hand, what the processor actually touches. The processor can only operate on registers. So everything, under the hood, will end up back there.
Now using interrupts correctly is a whole other topic. I again recommend the same book- Ch 5 and Appendix A7
Note- My assembly class was in MIPS. I've never specifically studied this microcontroller. If I get some of the architecture wrong, forgive me.
I am programming a PIC18F94K20 to work in conjunction with a MCP7941X I2C RTCC ship and a 24AA128 I2C CMOS Serial EEPROM device. Currently I have code which successfully intialises the seconds/days/etc values of the RTCC and starts the timer, toggling a LED upon the turnover of every second.
I am attempting to augment the code to read back the correct data for these values, however I am running into trouble when I try to account for the various 'extra' bits in the values. The memory map may help elucidate my problem somewhat:
Taking, for example, the hours column, or the 02h address. Bit 6 is set as 1 to toggle 12 hour time, adding 01000000 to the hours bit. I can read back the entire contents of the byte at this address, but I want to employ an if statement to detect whether 12 or 24 hour time is in place, and adjust accordingly. I'm not worried about the 10-hour bits, as I can calculate that easily enough with a BCD conversion loop (I think).
I earlier used the bitwise OR operator in C to augment the original hours data to 24. I initialised the hours in this particular case to 0x11, and set the 12 hour control bit which is 0x64. When setting the time:
WriteI2C(0x11|0x64);
which as you can see uses the bitwise OR.
When reading back the hours, how can I incorporate operators into my code to separate the superfluous bits from the actual time bits? I tried doing something like this:
current_seconds = ReadI2C();
current_seconds = ST & current_seconds;
but that completely ruins everything. It compiles, but the device gets 'stuck' on this sequence.
How do I separate the ST / AMPM / VBATEN bits from the actual data I need, and what would a good method be of implementing for loops for the various circumstances they present (e.g. reading back 12 hour time if bit 6 = 0 and 24 hour time if bit6 = 1, and so on).
I'm a bit of a C novice and this is my first foray into electronics so I really appreciate any help. Thanks.
To remove (zero) a bit, you can AND the value with a mask having all other bits set, i.e., the complement of the bits that you wish to zero, e.g.:
value_without_bit_6 = value & ~(1<<6);
To isolate a bit within an integer, you can AND the value with a mask having only those bits set. For checking flags this is all you need to do, e.g.,
if (value & (1<<6)) {
// bit 6 is set
} else {
// bit 6 is not set
}
To read the value of a small integer offset within a larger one, first isolate the bits, and then shift them right by the index of the lowest bit (to get the least significant bit into correct position), e.g.:
value_in_bits_4_and_5 = (value & ((1<<4)|(1<<5))) >> 4;
For more readable code, you should use constants or #defined macros to represent the various bit masks you need, e.g.:
#define BIT_VBAT_EN (1<<3)
if (value & BIT_VBAT_EN) {
// VBAT is enabled
}
Another way to do this is to use bitfields to define the organisation of bits, e.g.:
typedef union {
struct {
unsigned ones:4;
unsigned tens:3;
unsigned st:1;
} seconds;
uint8_t byte;
} seconds_register_t;
seconds_register_t sr;
sr.byte = READ_ADDRESS(0x00);
unsigned int seconds = sr.seconds.ones + sr.seconds.tens * 10;
A potential problem with bitfields is that the code generated by the compiler may be unpredictably large or inefficient, which is sometimes a concern with microcontrollers, but obviously it's nicer to read and write. (Another problem often cited is that the organisation of bit fields, e.g., endianness, is largely unspecified by the C standard and thus not guaranteed portable across compilers and platforms. However, it is my opinion that low-level development for microcontrollers tends to be inherently non-portable, so if you find the right bit layout I wouldn't consider using bitfields “wrong”, especially for hobbyist projects.)
Yet you can accomplish similarly readable syntax with macros; it's just the macro itself that is less readable:
#define GET_SECONDS(r) ( ((r) & 0x0F) + (((r) & 0x70) >> 4) * 10 )
uint8_t sr = READ_ADDRESS(0x00);
unsigned int seconds = GET_SECONDS(sr);
Regarding the bit masking itself, you are going to want to make a model of that memory map in your microcontroller. The simplest, cudest way to do that is to #define a number of bit masks, like this:
#define REG1_ST 0x80u
#define REG1_10_SECONDS 0x70u
#define REG1_SECONDS 0x0Fu
#define REG2_10_MINUTES 0x70u
...
And then when reading each byte, mask out the data you are interested in. For example:
bool st = (data & REG1_ST) != 0;
uint8_t ten_seconds = (data & REG1_10_SECONDS) >> 4;
uint8_t seconds = (data & REG1_SECONDS);
The important part is to minimize the amount of "magic numbers" in the source code.
Writing data:
reg1 = 0;
reg1 |= st ? REG1_ST : 0;
reg1 |= (ten_seconds << 4) & REG1_10_SECONDS;
reg1 |= seconds & REG1_SECONDS;
Please note that I left out the I2C communication of this.
How do I write to a single bit? I have a variable that is either a 1 or 0 and I want to write its value to a single bit in a 8-bit reg variable.
I know this will set a bit:
reg |= mask; // mask is (1 << pin)
And this will clear a bit:
reg &= ~mask; // mask is (1 << pin)
Is there a way for me to do this in one line of code, without having to determine if the value is high or low as the input?
Assuming value is 0 or 1:
REG = (REG & ~(1 << pin)) | (value << pin);
I use REG instead of register because as #KerrekSB pointed out in OP comments, register is a C keyword.
The idea here is we compute a value of REG with the specified bit cleared and then depending on value we set the bit.
Because you tagged this with embedded I think the best answer is:
if (set)
reg |= mask; // mask is (1 << pin)
else
reg &= ~mask; // mask is (1 << pin)
(which you can wrap in a macro or inline function). The reason being that embedded architectures like AVR have bit-set and bit-clear instructions and the cost of branching is not high compared to other instructions (as it is on a modern CPU with speculative execution). GCC can identify the idioms in that if statement and produce the right instructions. A more complex version (even if it's branchless when tested on modern x86) might not assemble to the best instructions on an embedded system.
The best way to know for sure is to disassemble the results. You don't have to be an expert (especially in embedded environments) to evaluate the results.
One overlooked feature of C is bit packing, which is great for embedded work. You can define a struct to access each bit individually.
typedef struct
{
unsigned char bit0 : 1;
unsigned char bit1 : 1;
unsigned char bit2 : 1;
unsigned char bit3 : 1;
unsigned char bit4 : 1;
unsigned char bit5 : 1;
unsigned char bit6 : 1;
unsigned char bit7 : 1;
} T_BitArray;
The : 1 tells the compiler that you only want each variable to be 1 bit long. And then just access the address that your variable reg sits on, cast it to your bit array and then access the bits individually.
((T_BitArray *)®)->bit1 = value;
® is the address of your variable. ((T_BitArray *)®) is the same address, but now the complier thinks of it as a T_BitArray address and ((T_BitArray *)®)->bit1 provides access to the second bit. Of course, it's best to use more descriptive names than bit1
//Through Macro we can do set resset Bit
#define set(a,n) a|=(1<<n);
#define reset(a,n) a&=(0<<n);
//toggle bit value given by the user
#define toggle(a,n) a^=(1<<n);
int a,n;
int main()
{
printf("Set Reset particular Bit given by User ");
scanf("%d %d",&a,&n);
int b =set(a,n) //same way we can call all the macro
printf("%d",b);
return 0;
}
I think what you're asking is if you can execute a write instruction on a single bit without first reading the byte that it's in. If so, then no, you can't do that. Has nothing to do with the C language, just microprocessors don't have instructions that address single bits. Even in raw machine code, if you want to set a bit you have to read the byte it's in, change the bit, then write it back. There's just no other way to do it.
Duplicate of how do you set, clear, and toggle a single bit and I'll repost my answer too as no-one's mentioned SET and CLEAR registers yet:
As this is tagged "embedded" I'll assume you're using a microcontroller. All of the above suggestions are valid & work (read-modify-write, unions, structs, etc.).
However, during a bout of oscilloscope-based debugging I was amazed to find that these methods have a considerable overhead in CPU cycles compared to writing a value directly to the micro's PORTnSET / PORTnCLEAR registers which makes a real difference where there are tight loops / high-frequency ISR's toggling pins.
For those unfamiliar: In my example, the micro has a general pin-state register PORTn which reflects the output pins, so doing PORTn |= BIT_TO_SET results in a read-modify-write to that register.
However, the PORTnSET / PORTnCLEAR registers take a '1' to mean "please make this bit 1" (SET) or "please make this bit zero" (CLEAR) and a '0' to mean "leave the pin alone". so, you end up with two port addresses depending whether you're setting or clearing the bit (not always convenient) but a much faster reaction and smaller assembled code.