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I know this may be wrong section for this but my problem is Microcontroller programming specific (AVR mostly)!

I am sending bytes between two AVR atmega8 using Uart where each bit in the byte stands for something and only one bit is 1 in each byte sent

So if I want to check, say for example, 5th bit in the received byte then if write it as follows:

short byte=UDR;
if(byte&(1<<5))
{
// do stuff for bit 5
}

Then it works fine always

BUT, if I write it as this:

short byte=UDR;
if(byte==0b00100000)

OR
short byte=UDR;
if(byte==0x20)

Then it won't work , also it fails if I use Switch-case instead of if-else I can't understand the problem, does the compiler interprets it as signed no and 7th Bit as sign? or something else? The compiler is AVR-gnu from AVR studio 5

If someone asks I also have LEDs on the receiver that shows received byte and so I know the byte received is correct but for some reason I cannot compare it for conditions! still can there some noise that causes bits to be misunderstood by the Uart and thus changing the actual byte received?

Help !

LOOK OUT EVERY ONE

Something here is like PARANORMAL

Finally I have cornered the area of problem

I added 8 LEDs to represent the bits of received bytes and here's what I found:

The LEDs represent (1<<5) as 0b00100000 that's ok as its what I sent

BUT

The other LEDs (excluding the 8) assigned to glow on receiving 0b00100000 does not glow!

FTW MAN!

I'm damn sure the received byte is correct ..but something's wrong with if-else and switch-case comparision

user1448559
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2 Answers2

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C doesn't have a syntax for binary literals, you can't type 0b00100000 and have it compile.

It's a bit hard to understand why it wouldn't work for the == 0x20 case, since we don't know the value of UDR that is specific to your platform.

If UDR has more than 1 bit set, then the exact equality check will of course fail, while the single bit test will succeed.

You can only use switch() with exact values for each case, but you can of course mask before inspecting:

switch( byte & 0x20 )
{
  case 0:
   break;
  case 0x20:
   break;
}
unwind
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  • The standard version of C, i.e. what is generally meant when talking about C. And no, C++ also doesn't have binary integer literals. – unwind May 04 '15 at 08:40
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    Hmm, I guess it's a feature of some compilers. I have used 0b00000000 as a binary representation of an array of sensors once. It did work great, even in a `switch-case`. I found out that C++14 does support it though. http://en.wikipedia.org/wiki/C%2B%2B14#Binary_literals – Paul May 04 '15 at 08:52
  • @FuaZe Good catch, I wasn't aware of that. And yes, of course compilers can provide extensions. Sometimes a programming environment can also provide symbolic constants that look similiar, such at the (semi-horrible, in my opinion) [`binary.h`](https://code.google.com/p/arduino/source/browse/trunk/hardware/arduino/cores/arduino/binary.h) used with the Arduino. – unwind May 04 '15 at 09:03
  • https://gcc.gnu.org/onlinedocs/gcc/Binary-constants.html Indeed, found the documentation of Gcc GNU, where it specifies the binary literals. Though, arduino is based on GCC GNU? It makes me wonder why they use such an (semi-horrible indeed) option. – Paul May 04 '15 at 09:08
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It doesn't work because the second formulation changes the meaning of the code, not just the spelling of the mask constant. To test a bit, you must apply the bitwise and (&) to the constant, not just compare the value with the constant:

if (byte & 0b00100000)    /* note: 0b00100000 is a gcc extension */

or:

if (byte & 0x20)          /* or byte & 32, or byte & (1 << 5), etc. */
user4815162342
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  • NOOOOO!!! that's not what I'm saying ... re-read the question there is only one bit that's high so if(byte==0b00100000) must be equivalent to if(byte&(1<<5)) – user1448559 Jan 14 '13 at 14:38
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    @user1448559 Well, that's obviously not the case, since the first version works, and the second doesn't. If you're not sure, simply print the received value. – user4815162342 Jan 14 '13 at 14:50
  • I Agree with user4815162342. & is the way to test for bits. You can check if only one bit is really set by Xor-ing with 0x20 and checking for 0 – deStrangis Jan 14 '13 at 15:11
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    @deStrangis XOR-ing with the mask and checking for 0 works, but it's a roundabout way of spelling `byte == 0x20`. Since the OP already tried that and it failed, one must conclude that other bits are also being set. – user4815162342 Jan 14 '13 at 16:12