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

[alexan_e]

You said in a previous post that you want to check which bits become 0, you can go to the basic form of doing that

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Code C - [expand]
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if !CHECKBIT(PORTX,0) ...;   // true if bit0 is 0
if !CHECKBIT(PORTX,1) ...;
if !CHECKBIT(PORTX,2) ...;
if !CHECKBIT(PORTX,3) ...;
if !CHECKBIT(PORTX,4) ...;
if !CHECKBIT(PORTX,5) ...;
if !CHECKBIT(PORTX,6) ...;
if !CHECKBIT(PORTX,7) ...;

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But what do you want to do with the result of each bit?

Alex

 

[romel_emperado]

hi alexan,,


actually this is what i want.. this is my finished code..
i also attached the protues file and the hex. now its totally done.. hehe

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Code C - [expand]
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#define SETBIT(ADDRESS,BIT) (ADDRESS |= (1<<BIT))
#define CHECKBIT(ADDRESS,BIT) (ADDRESS & (1<<BIT))
void main()
{
  char i;
  P3=0x00;
  P1=0xFF;
 
  while(1)
  {
 
      if (CHECKBIT(~P1,0))
      {
         P3= 1;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,1))
      {
         P3= 2;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,2))
      {
         P3= 3;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,3))
      {
         P3= 4;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,4))
      {
         P3= 5;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,5))
      {
         P3= 6;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,6))
      {
         P3= 7;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,7))
      {
         P3= 8;
      }
      Delay_ms(300);
      if (CHECKBIT(~P1,8))
      {
         P3= 9;
      }
      else P3=0;
      Delay_ms(300);
      
 
     }
  }

---

 

[raco_rage]

Tell me how the code above and this code behave differently??

void main()
{
  P3=0x00;
  P1=0xFF;

  while(1)
  {
    for(i = 0; i<8; i++)
    {
      if(~P1 & (0x01 << i))
        P3 = i + 1;
     else
       P3 = 0;
     Delay_ms(300);
    }
  }
}

 

[alexan_e]

Actually your code is different from what he does because for each i value you either set the output to a value or turn off everything,
in his code he only assigns value to the output after the check of each bit and he only clears the output port using if/else for bit 7 (you use it for all bits).
I know that you can easily change it to have exactly the same behavior, other than that the for loop is a more refined way.
That being said i think that both codes don't behave as they should because the output should be 0 only if all the input bits are 1 so it would reasonable to use a check after the loop ends and if the port is 0xff then turn everything off.

Alex

 

[raco_rage]

Thanks alexan_e.

So, this code is what the OP really wanted,

void main()
{
  P3=0x00;
  P1=0xFF;

  while(1)
  {
    for(i = 0; i<8; i++)
    {
      if(~P1 & (0x01 << i))
        P3 += i + 1;
      if(~P1 == 0)
        P3 = 0;
     Delay_ms(300);
    }
  }
}

 

[alexan_e]

I'm not sure what he wants but this resembles the functionality of the code he wrote
with the addition of the clear output when input is 0xff
as i think it should be but maybe he doesn't want it this way.

The only other change i would make would be the delay,
I don't understand the purpose of a delay between each step (but that is what he did),
i would probably use the delay out of the loop and then add if(P1 == 0xff) P3 = 0; before the loop executes again.

Alex

 

[romel_emperado]

hi raco, your last posted code is what i am using now.... it's very nice and efficient..

hi alexan, I added delay because i just want to delay and see the simulation properly....
as what i have said, my programming skill is not good enough to create good algorithm for my programs. I only knew basic style of codding.. as of now, it is okay to me if i see my code working as it should be but i dont know hot to write a code efficiently like what raco did to my code..

yesterday I am tired of using a very long switch cases just to get all of the combination of 8bits and my code exceeds 2kb of memory which is not enough if i will use AT89c2051.


it is so funny to think about yesterday this was my code of getting the low bits... :-D:-D

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Code C - [expand]
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void main()
{
 while(1)
 {
   switch(P1)
   {
       case 0:
       case 1:
       .
       .  
       . 
       .
       case 255:
 
 
   }
 
 }
}

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and then after now the equivalent to my long cases of switch is just like this.. very short compared to my style... :-D:-D Im very thankful guys... thanks for helping

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Code C - [expand]
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void main()
{
  char i;
  P3=0x00;
  P1=0xFF;
 
  while(1)
  {
    for(i = 0; i<8; i++)
    {
      if(~P1 & (0x01 << i))
       {
        Delay_ms(100);
        P3 += i + 1;
        Delay_ms(100);
       }
      if(~P1 == 0)
        Delay_ms(100);
        P3 = 0;
 
    }
 
  }
}

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---------- Post added at 01:51 ---------- Previous post was at 01:44 ----------

thanks all of you guys for contributing my learning

 

[alexan_e]

hi alexan, I added delay because i just want to delay and see the simulation properly....

actually i didn't say to remove the delay, i said to reposition it.
The delay is used in the above code in each step of the loop, this means that if you provide a 0 to bit7 you will have to wait for 7*200ms for that to be detected.
What i propose is this

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Code C - [expand]
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void main()
{
  char i;
  P3=0x00;
  P1=0xFF;
 
  while(1)
  {
    for(i = 0; i<8; i++)
    {
      if(~P1 & (0x01 << i))
       {
         P3 = i + 1;
        // break;  // you can also add a break here to exit the loop when an input is detected
       }
    }
    
    if (P3 != 0)  // if the output is 0 no reason to delay or clear
    {
        Delay_ms(500);   //delay
        P3 = 0;  // and then clear everything
    }
  }
}

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using this code you will read immediately the input when the output is off
and the output will hold the value for the delay time

I think the line P3 += i + 1; should be P3 = i + 1; as it was in the previous post so that the output is 1 to 8 depending on the input, id depends what you want.

Alex

 

[romel_emperado]

thanks alexan.... I will apply that...

 

[romel_emperado]

i will mark this thread as solved now.. thanks guys..