The time has come...

**budgie** · 01-09-2005

First of all id like to thank cprogramming.com and its member's for its fantastic tutorials and faq's as between you guy's and google you've taught me all i know about c/c++

ive been to writing a simple program to control a parallel port I/O card, but the time has come where i need a little help.

the code snip below is what im using to check the variable "input" against the input_address array, this works fine for checking 1 input line at a time but if 2 or more inputs are pulled high it dosent work.

let say input 1 and 2 are high the variable "input" will = 3
or if all 8 are high "input" will = 255
i need it to print out, inputs 1 and 2 are high or inputs 1, 2, 3, 4, 5, 6, 7 and 8 are high and whatever in between e.g.,

if the variable "input" == 82 then print inputs 2, 5 and 7 are high

Code:

int input_adress[10] = {0,1,2,4,8,16,32,64,128};

for (i = 1;i < 9;i++)
  { if (input == input_adress[i])
    {
      cout << "Input " << i << " is HIGH" << endl; }
    else 
    { input == 0;
     cout << "Input " << i << " in LOW" << endl; }}

and for my second question, in the line below how do i put these : every 2 digits to get the output to look like this 12:01:01 instead of 120101

Code:

cout << "The time is"<< setfill('0') << setw(6) << thetime << endl;

Thank you for your very much time Budgie

**Prelude** · 01-09-2005

That's somewhat of a tricky formatting scheme. If you do it with straight printing, it would end up looking like this:

Code:

#include <iostream>

using namespace std;

int main()
{
  int input_address[10] = {0,1,2,4,8,16,32,64,128};
  int input = 4;
  bool high = false;

  for (int i = 1; i < 9; i++)
  {
    if (input > input_address[i])
    {
      high = true;

      // Not the first item
      if (i != 1) {
        if (i == 8 || (i + 1 < 9 && input == input_address[i + 1]))
          cout << " and ";
        else
          cout << ", ";
      }
      cout << i;
    }
  }
  if (!high)
    cout << "No items";
  cout << " are high" << endl;
}

Not very pretty. A better way is to save the high values and then pretty print them after the processing is done:

Code:

#include <iostream>

using namespace std;

int main()
{
  int input_address[10] = {0,1,2,4,8,16,32,64,128};
  int high[8] = {0};
  int n = 0;
  int input = 4;

  for (int i = 1; i < 9; i++)
  {
    if (input > input_address[i])
      high[n++] = i;
  }
  
  if (n == 0)
  {
    cout << "No items are high" << endl;
  }
  else {
    cout << (n == 1 ? "Input " : "Inputs ");
    for (int i = 0; i < n; i++)
    {
      cout << high[i];
      if (i + 1 == n - 1)
        cout << " and ";
      else if (i != n - 1)
        cout << ", ";
    }
    cout << (n == 1 ? " is " : " are ") << "high" << endl;
  }
}

It might not be as concise, but it's easier to get right, and localizing/separating operations will greatly simplify your code. The nice part is that now you can modularize the pretty print into a function:

Code:

#include <iostream>

using namespace std;

void print(int high[], int n)
{
  if (n == 0)
  {
    cout << "No items are high" << endl;
  }
  else {
    cout << (n == 1 ? "Input " : "Inputs ");
    for (int i = 0; i < n; i++)
    {
      cout << high[i];
      if (i + 1 == n - 1)
        cout << " and ";
      else if (i != n - 1)
        cout << ", ";
    }
    cout << (n == 1 ? " is " : " are ") << "high" << endl;
  }
}

int main()
{
  int input_address[10] = {0,1,2,4,8,16,32,64,128};
  int high[8] = {0};
  int n = 0;
  int input = 4;

  for (int i = 1; i < 9; i++)
  {
    if (input > input_address[i])
      high[n++] = i;
  }

  print(high, n);
}

>every 2 digits to get the output to look like this 12:01:01 instead of 120101
Is thetime and integer or a string? If it's a string then the task is a simple matter of printing two characters then a colon:

Code:

#include <iostream>

using namespace std;

int main()
{
  char thetime[] = "120101";

  cout << "The time is "
    << thetime[0] << thetime[1] << ':'
    << thetime[2] << thetime[3] << ':'
    << &thetime[4] << endl;
}

C++ strings make it much cleaner:

Code:

#include <iostream>
#include <string>

using namespace std;

int main()
{
  string thetime = "120101";

  cout << "The time is "
    << thetime.substr(0, 2) << ':'
    << thetime.substr(2, 2) << ':'
    << thetime.substr(4) << endl;
}

Though if you're using strings, you should have formatted them correctly to begin with, so I suspect you're using an integer. That's harder, but because times are well formatted, you can do something like this:

Code:

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{
  int thetime = 120101;

  cout << "The time is "
    << setw(2) << setfill('0') << thetime / 10000 << ':'
    << setw(2) << setfill('0') << (thetime / 100) % 10 << ':'
    << setw(2) << setfill('0') << thetime % 10 << endl;
}

Too bad it's not a good one-liner.

**budgie** · 01-09-2005

wow thanks for the fast replys, its gunna take me a bit to get my head around this, i'll get back to you guy's soon as i do

For the second question: what is the exact nature of the variable "thetime"?

If its value is decimal 120101, does this indicate one second past one minute past noon, or what? (Is this on a 24-hour clock?)

yeah it does indicate one second past one minute past noon

its an integer of a 24 hour clock

Thanks for your time, budgie

**sean** · 01-09-2005

wow thanks for the fast replys

55 minutes is a slow response time here, actually. But not for THAT kind of response.

**Dave Evans** · 01-09-2005

Originally Posted by Prelude

Though if you're using strings, you should have formatted them correctly to begin with, so I suspect you're using an integer. That's harder, but because times are well formatted, you can do something like this:

Code:

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{
  int thetime = 120101;

  cout << "The time is "
    << setw(2) << setfill('0') << thetime / 10000 << ':'
    << setw(2) << setfill('0') << (thetime / 100) % 10 << ':'
    << setw(2) << setfill('0') << thetime % 10 << endl;
}

Not quite (try it with thetime =123456).

You could do this:

Code:

    hours = thetime/10000;
    minutes = (thetime - hours*10000)/100;
    seconds = thetime - hours*10000 - minutes*100;

Regards,

Dave

**Prelude** · 01-09-2005

>Not quite (try it with thetime =123456).
Oops, my 10's should have been 100's. Serves me right for not compiling and testing my own example.

**budgie** · 01-09-2005

Thank's guys the time formating work's well
i wrote a small program to test the input stuff, its close but dosent retern the right values

Code:

#include <iostream>
#include <sys/io.h>

using namespace std;

#define PORTADDRESS 0x378  		 // Parallel port address
#define DATA PORTADDRESS	         // DATA address
#define STATUS PORTADDRESS+1	         // STATUS address
#define CONTROL PORTADDRESS+2		 // CONTROL address

int main()
{
  int card_on, input, i;
  int input_address[10] = {0,1,2,4,8,16,32,64,128};
  int cards[10] = {0,10,8,14,12,2,0,6,4};
  int low[8] = {0};
  int n = 0;
  //int input = 4; took this out

  cout <<"Select what card to use ( 1-8 ) :> ";
  cin >> card_on;
  cout << "Using Card Number " << card_on << endl ;

  if (ioperm(PORTADDRESS,3,1)) {perror("ioperm");exit(1);} // Open the parllel port

  outb(cards[card_on], CONTROL);    // Select card
  outb(cards[card_on]+1, CONTROL);  // Latch data

  for(i = 1; i <= 128; i++)
  {
    outb(120, STATUS);            
    outb(5, CONTROL);          
  }
  input = (inb(STATUS) & 248) ^ 120;            
  input =  input | ((inb(CONTROL) & 14) ^ 4) / 2;  
  outb(cards[card_on], CONTROL);                 
  cout << "the input value is "<< input << endl;

  // your example
  for (int i = 1; i < 9; i++)
  {
    if (input > input_address[i])
      low[n++] = i;
  }

  if (n == 0)
  {
    cout << "No items are low" << endl;
  }
  else
  {
    cout << (n == 1 ? "Input " : "Inputs ");
    for (int i = 0; i < n; i++)
    {
      cout << low[i];
      if (i + 1 == n - 1)
        cout << " and ";
      else if (i != n - 1)
        cout << ", ";
    }
    cout << (n == 1 ? " is " : " are ") << "low" << endl;
  }
  // end example

if (ioperm(PORTADDRESS,3,0)) {perror("ioperm");exit(1);}  // Close the parllel port

  exit(0);
}

here's the output, the first one is working input = 0, all inputs are high (had my highs and lows the wrong way around in my fist post)
the second one input value = 4 should be input 3 is low
the third value = 254 should be inputs 2,3,4,5,6,7 and 8 are low

Code:

[root@home src]# ./jaycar_testbed
Select what card to use ( 1-8 ) :> 1
Using Card Number 1
input value is 0
No items are low

[root@home src]# ./jaycar_testbed
Select what card to use ( 1-8 ) :> 1
Using Card Number 1
input value is 4
Inputs 1 and 2 are low

[root@home src]# ./jaycar_testbed
Select what card to use ( 1-8 ) :> 1
Using Card Number 1
input value is 254
Inputs 1, 2, 3, 4, 5, 6, 7 and 8 are low
[root@home src]#

i did take this out "int input = 4;" and replaced it with the integer reterned from the I/O card is this the prob.

once again thanks for your time

**Prelude** · 01-09-2005

You're making less sense now.

>the second one input value = 4 should be input 3 is low
Okay, but what about 1 and 2? And item 3 is 4, which is equal to input. Is that considered high or low?

>the third value = 254 should be inputs 2,3,4,5,6,7 and 8 are low
What about 1? I don't see a pattern in these new results, and it looks like you want something completely different from what you asked for previously.

**budgie** · 01-09-2005

sorry im not to good at explaning things, i'll try again

the variable "input" is going to be a number somewhere between 0 and 255 depending on what state the card inputs r in.

so if the input == 254 in binary it = 11111110

76543210
. . . . . . . . 1
. . . . . . * . 2
. . . . . * . . 4
. . . . * . . . 8
. . . * . . . . 16
. . * . . . . . 32
. * . . . . . . 64
* . . . . . . . 128

so 2 + 4 + 8 +16 +32 + 64 + 128 = 254 so inputs 2,3,4,5,6,7,and 8 are low

so if the input == 48 in binary it = 00110000

76543210
. . . . . . . . 1
. . . . . . . . 2
. . . . . . . . 4
. . . . . . . . 8
. . . * . . . . 16
. . * . . . . . 32
. . . . . . . . 64
. . . . . . . . 128

16 + 32 = 48 inputs 5 and 6 are low

can you see what i mean, i should realy number the inputs 0-7 not 1-8

**Dave Evans** · 01-09-2005

Originally Posted by budgie

sorry im not to good at explaning things, i'll try again

the variable "input" is going to be a number somewhere between 0 and 255 depending on what state the card inputs r in.

so if the input == 254 in binary it = 11111110

76543210
. . . . . . . . 1
. . . . . . * . 2
. . . . . * . . 4
. . . . * . . . 8
. . . * . . . . 16
. . * . . . . . 32
. * . . . . . . 64
* . . . . . . . 128

so 2 + 4 + 8 +16 +32 + 64 + 128 = 254 so inputs 2,3,4,5,6,7,and 8 are low

so if the input == 48 in binary it = 00110000

76543210
. . . . . . . . 1
. . . . . . . . 2
. . . . . . . . 4
. . . . . . . . 8
. . . * . . . . 16
. . * . . . . . 32
. . . . . . . . 64
. . . . . . . . 128

16 + 32 = 48 inputs 5 and 6 are low

can you see what i mean, i should realy number the inputs 0-7 not 1-8

The way that you are testing the bits is fundamentally wrong.

Why not use a bit mask?

Test this with whatever inputs you want to try, then if it looks OK, you can integrate it into your program to get the output formatted as required.

Code:

#include <iostream>
using namespace std;

int main()
{
  int input;
  int mask;
  int i;

  while (1) {
    cout << "Enter a number: " << flush;
    cin >> input;
    if (!cin) {
      break;
    }
    cout << "You entered " << input << endl;

    mask = 1; // start '1' in lsb
    for (i = 0; i < 8; i++) // number lsb = 0, msb = 7
    {
      if ((input & mask)) { // check for '1'
        cout << "  bit " << i << " is high" << endl;
      }
      mask <<= 1;
    }
    cout << endl;
  }
  return 0;
}

Regards,

Dave

**budgie** · 01-09-2005

you the man Dave, thank you very much, that's what i need.

Prelude thanks for your help m8, sorry about confusion

all the best, budgie

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