[SOLVED] basic question about c

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syenidogan

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if ((buffer[1] == 16)&(((unsigned int) buffer[6] + 9)!=counter))


(buffer[1]==16) gives one bit result 1 or 0
(((unsigned int) buffer[6] + 9)!=counter) gives one bit result 1 or 0

then and (&) these two values and see if it is 1 or zero ? is this correct ?


if not ,can some one explain me detailed.

buffer is unsigned char
and counter is unsigned int
 
Last edited:

What type of variables are counter, and buffer[] ?

Here it is checked if buffer[1] equals 16 and buffer[6] casted to unsigned int + 9 not equal to value in counter variable.
 

milan.rajik said:
What type of variables are counter, and buffer[] ?

Here it is checked if buffer[1] equals 16 and buffer[6] casted to unsigned int + 9 not equal to value in counter variable.
Click to expand...

buffer is unsigned char
and counter is unsigned int
 

If buffer[1] equals to 16 then it gives 1 (true) and if not equal it gives 0 (false)

Then buffer[6] is casted to unsigned int. If buffer[6] which is unsigned char (unsigned char can hold 256 values, 0-255) then it is casted to unsigned int which will also have the same value. I don't know why casting is done here. It doesn't make any difference. Then 9 is added to this value and it is compared whether this value is not equal to value in counter. If it is not equal you get 1 (true) and you get 0 (false) if it is equal to counter. Then if you get both trues then the if condition is executed.

I think && is typed as & in your code.
 


if ((buffer[1] == 16)&(((unsigned int) buffer[6] + 9)!=counter)) is it the same thing
if ((buffer[1] == 16)&&(((unsigned int) buffer[6] + 9)!=counter))

if not which one is correct? if both of them correct what is the difference?
 

I think he is asking if its a problem to use a bitwise AND (&) in place of a logical AND (&&).
It may work in some cases but you are better off using && which is the proper one for logic operations.

Suppose that you want to check if two bits are high

Code: 
if( (PINB & (1<<1)) & (PINC & (1<<2)) )

Assume that this equals

Code: 
 if( (0x02)) & (0x04)) )

this would give a false result because ((0x2) & (0x04)) results to 0

Now if this was written with &&

Code: 
 if( (0x02)) && (0x04)) )

then it would be the equivalent of

Code: 
if( (1)) && (1)) )

and the result would be true

&& between non 0 values results to 1.
 

Maybe this is the correct one

Code C - [expand]
1

if( ( (buffer[1] == 16) && ((unsigned int) buffer[6] + 9)) ) != counter)
 


this means that (buffer[1]==16) if it gives zero and (((unsigned int) buffer[6] + 9)!=counter) if this also gives zero

then if do & i will get zero but if i do && i will get 1 ?
 

I am extremely sorry. This is the correct one.

Code C - [expand]
1

if ((buffer[1] == 16)&&(((unsigned int) buffer[6] + 9)!=counter))


without && it will give error. If any one is 0 the if will not execute. if will execute if only both are 1 that is true.

true && true = true
true && false = false
false && true = false
false && false = false
 

syenidogan said:
this means that (buffer[1]==16) if it gives zero and (((unsigned int) buffer[6] + 9)!=counter) if this also gives zero

then if do & i will get zero but if i do && i will get 1 ?
Click to expand...

No

Code: 
0x00 & 0x00 results to 0
0x01 & 0x00 results to 0
0x01 & 0x01 results to 1
0x01 & 0x02 results to 0
0x02 & 0x02 results to 2

0x00 & 0x00 results to 0
0x01 & 0x00 results to 1
0x01 & 0x01 results to 1
0x01 & 0x02 results to 1
0x02 & 0x02 results to 1


&& between 0 values results to 0.
&& between non 0 values results to 1.
 


confused!!!! (

- - - Updated - - -

just tell me which values should be in buffer[1] and buffer[6] if my counter is 15 to do execute return?

if ((buffer[1] == 16)&(((unsigned int) buffer[6] + 9)!=counter)) return 2 ;

buffer[1]=0x10 and buffer[6] different than 0x06 ?
 

All I'm saying is that you are using the bitwise AND operator in place of a logical AND operator. In your example both will give a correct result but the proper one is the && operator since you want to logically AND the two conditions.
 

If buffer[1] contains 16 then 
Code C - [expand]
1

(buffer[1] == 16)
becomes true (1) and if counter is 15 then counter - 9 = 15 - 9 = 6. So, buffer[6] should contain 6. If buffer[1] = 16 and buffer[6] = 6 then both conditions become true (1) and hence if condition will execute.
 

Yes.

If buffer[1] contains 16 then (buffer[1] == 16)
becomes true (1) and if counter is 15 then counter - 9 = 15 - 9 = 6. So, buffer[6] should not contain 6. If buffer[1] = 16 and buffer[6] = 6 then both conditions become true (1) and hence if condition will execute.
 

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