[SOLVED] help me in programming :) detecting the input LOW bits..

[romel_emperado]

guys, do you have any hint on how to program this?

this is the problem.
example if i have input in POR1 = 1111 1010

i need to know which bit is low. as seen in the example there are two low/0 values in the input... how to detect that??

actually from the example P1.0 and P1.2 are low.


i can do that using switch case but i think there is another method to do that. imagine that is 8bits i cant manage to put all that case combination by using switch case from 0 - 255. :sad:

 

[raco_rage]

simply run in a for loop,

example,

for(i = 0; i<=8; i++)
{
if(POR1 & (0x01 << i))
BIT IS SET
else
BIT CLEAR
}

 

[romel_emperado]

hi raco, what do you mean by bit is set and bit is clear?

how does it actually work?

as i understand it is scanning from P1.0 to P1.7

but i dont actually get the application of bit is set? its my first time to know about that :sad:

im thinking to output the data to another port.. let say if i detected that P1.0 is low then it will output as 1 then if the last bit(P1.7) is low the output is 8..

 

[raco_rage]

Bit set and clear is another term for saying that bit is 1 and 0, respectively.

It becomes even easier if you want to output to another port, simply output the value of "i" when you detect a high bit.


for(i = 0; i<8; i++)
{
if(POR1 & (0x01 << i))
PORTX = i + 1;
}

 

[romel_emperado]

i dont realy get it.. i tried it using proteus but it only counts 1 to 8 ;(

this line if(P1 & (0x01 << i)) dont take it's effect

for(i = 0; i<8; i++)
{
Delay_ms(2000);
if(P1 & (0x01 << i))
P3 = i + 1;
}

 

[raco_rage]

so probably you could try,

for(i=1; i<256; i= i*2)
if(P1 & i)
P3 = i;

But logically, they both should work fine and are same.

 

[alexan_e]

for(i = 0; i<8; i++)
{
if(POR1 & (0x01 << i))
PORTX = i + 1;
}

If you want to set the bit of the other port then PORTX = i + 1; will not do much, it can only take values from 1 to 8.
If you want to set or clear the bits of the other port then you should use

---

Code C - [expand]
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for(i = 0; i<8; i++)
{
    if(POR1 & (0x01 << i)) 
        PORTX |= (1<<i); // set bit
    else 
        PORTX &= ~(1<<i);   // clear bit
}

---

but if you just want to copy a port to another port then why not use PORY=PORX;

Alex

 

[Klen]

Both of the replies from raco_rage's posts are spot on and totally correct. However, if I may add, romel_emperado mentioned that he wants to output a numerical value everytime he encounters a low bit. The following code (actually 99.99% replica of raco_rage's code) should work. If not, then I believe the problem should be somewhere else in the code.

for(i = 0; i<8; i++)
{
if(!(POR1 & (0x01 << i))) // TRUE for bit = 0
PORTX = i + 1;
}

(Note: Are you sure you want to output an integer to the port at ever occurence of a corresponding low bit? Normally, one would have to bitmask a byte on to a port.....)

 

[romel_emperado]

(Note: Are you sure you want to output an integer to the port at ever occurence of a corresponding low bit? Normally, one would have to bitmask a byte on to a port.....)

yes i that exactly i want to do. i need to detect what pins of P1 is low or high.....


guys, thanks for your help.. i will try your suggestions...

 

[raco_rage]

yea, exactly as Klen says, there should be a ! for low.

As the OP says that he wants to output values only from 1 to 8, corresponding to each bit position, he wouldn't need a output value more than 8 unless he is checking a 16 or 32 bit number.

 

[alexan_e]

I'm still not clear what exactly you are trying to do,
do you want to have

input out
00000001 -> 00000001
00000010 -> 00000010
00000100 -> 00000011
00001000 -> 00000100
00010000 -> 00000101

please give an example of the input and output you expect.

Alex

 

[romel_emperado]

please give an example of the input and output you expect.

example if i have this input:

P1 = 1011 1110

i must know that the two zeros in P1 are located in PIN1 of port1 and the other zero is in port7 of port1... that's what i want to do.. i dont have idea how to do that.. ;(

 

[Klen]

example if i have this input:

P1 = 1011 1110

i must know that the two zeros in P1 are located in PIN1 of port1 and the other zero is in port7 of port1... that's what i want to do.. i dont have idea how to do that.. ;(

The easiest way I can think of right now is to light some LEDs in active low logic. The code is already provided so it shouldnt be difficult

 

[raco_rage]

do u mean you want,

P3 = ~P1 ??

 

[romel_emperado]

do u mean you want,

P3 = ~P1 ??

nope.. sorry for causing you trouble guys..

this is the problem.. my port1 is monitoring for possibe input.. port1 has 8bits and then for example if P1 detected a value of 253. 253 in binary is 1111 1101

now, i want to know which PIN is low.... obviously if we look at the example input the low bit is in P1.1. let say for example i want to display like this "the zero value of 253 is located at P1.1"

something like that guys..


here is an example code i am using but it's not perfect it cant detect the whole 8bits..

---

Code C - [expand]
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int x = 10100101;
// First bit
(x >> 0) % 2; // 1 is detected that it is in  P1.0
// Second bit
(x >> 1) % 2; // 0 is detected that it is in  P1.1
// Third bit
(x >> 2) % 2; // 1 is detected that it is in  P1.2
...
etc.

---

 

[alexan_e]

There are some useful defines that you can use , just put them in the start of your code and you can use them after that

---

Code C - [expand]
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#define SETBIT(ADDRESS,BIT) (ADDRESS |= (1<<BIT)) 
#define CLEARBIT(ADDRESS,BIT) (ADDRESS &= ~(1<<BIT)) 
#define FLIPBIT(ADDRESS,BIT) (ADDRESS ^= (1<<BIT)) 
#define CHECKBIT(ADDRESS,BIT) (ADDRESS & (1<<BIT)) 
#define WRITEBIT(RADDRESS,RBIT,WADDRESS,WBIT) (CHECKBIT(RADDRESS,RBIT) ? SETBIT(WADDRESS,WBIT) : CLEARBIT(WADDRESS,WBIT))

---

To use them

---

Code C - [expand]
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SETBIT(PORTX,2);  // set bit2 of portx (set to 1)
SETBIT(PORTY,0);  // set bit0 of porty
 
CLEARBIT(PORTX,5); // clear bit5 of portx (set to 0)
 
FLIPBIT(PORTX,2);  // invert bit2 of portX, if it was 1 it becomes 0, and if 0 becomes 1
                    // example for portx 0x10101010 the result is 0x10101110
                    
if (CHECKBIT(PORTX,4)) {} else {} // result is true if bit4 of portx is 1, false if it is 0
 
WRITEBIT(PORTX,0,PORTY,2);   // copy the bit0 of portx to bit2 of porty

---

Alex

 

[romel_emperado]

hi alexan_e,

thanks for that.. i will surely try that...

---------- Post added at 16:54 ---------- Previous post was at 16:48 ----------

thanks alexan_e,

it solves my problem.. excellent!!! :-D:-D:-D:-D

this is my code actually

---

Code C (Mac) - [expand]
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#define SETBIT(ADDRESS,BIT) (ADDRESS |= (1<<BIT))
void main()
{
  SETBIT(P1,2);
  P3 = SETBIT(P1,0);
}

---

---------- Post added at 16:59 ---------- Previous post was at 16:54 ----------

to all of your.. thank you so much and sorry for causing your trouble

 

[alexan_e]

SETBIT(P1,2); sets P1 bit2 to 1

SETBIT(P1,0); sets P1 bit0 to 1 and copies the result byte to P3, is this what you want?

Alex

 

[raco_rage]

even if he was able to solve the problem, I am still not clear what he really wanted?

 

[romel_emperado]

SETBIT(P1,2); sets P1 bit2 to 1

SETBIT(P1,0); sets P1 bit0 to 1 and copies the result byte to P3, is this what you want?

Alex

yes that's really what i want.. i million thanks to you

actually the logic behind this solves my problem

#define SETBIT(ADDRESS,BIT) (ADDRESS |= (1<<BIT))

---------- Post added at 17:17 ---------- Previous post was at 17:09 ----------

oops sorry raco_rage , actually its not totally solved. im thinking of another but alexan_e give me idea..

its hard to explain how i solve my problem due to trial and error... hehe Im new to programming and Im learning just myself relying on some tut found in google.. thanks for helping guys