ZuoTi.Pro
Question

C Language: Write a diagnostic code for the following code. And please tell me what it...

C Language: Write a diagnostic code for the following code. And please tell me what it will check for. This code does the following: its using 2 7 Segment LEDs

Hit Switch

Count Up

Hit Switch

Display Time

Hit Switch

Count Down from 99

Hit Switch

Display time

Hit Switch

Turn off display

The diagonistic code should check: All segement of the LEDs and or buttons to make sure they work fine.

#include
#include
#include
#include
#include "LabJackUD.h"
#include "LJUD_DynamicLinking.h"


void ErrorHandler (LJ_ERROR ljError, long lngLineNumber)
{
    char err[255];
    if (ljError != LJE_NOERROR)
    {
        ErrorToString(ljError, err);
        printf("Error # %ld: %s\n", ljError, err);
        printf("Source line number = %ld\n", lngLineNumber);
        if(ljError > LJE_MIN_GROUP_ERROR)
        {
            getchar();
            exit(0); // Quit if serious error
        }
    }
}


const int numbers[10][1] =
{
    // 1: segment lit, 0: segment unlit
    //           a b c d e f g
    {64},   //0 1, 1, 1, 1, 1, 1, 0 1000000
    {121}, //1 0, 1, 1, 0, 0, 0, 0 1111001
    {36},   //2 1, 1, 0, 1, 1, 0, 1 0100100
    {48},   //3 1, 1, 1, 1, 0, 0, 1 0110000
    {25},   //4 0, 1, 1, 0, 0, 1, 1 0011001
    {18},   //5 1, 0, 1, 1, 0, 1, 1 0010010
    {2},    //6 1, 0, 1, 1, 1, 1, 1 0000010
    {120}, //7 1, 1, 1, 0, 0, 0, 0 1111000
    {0},    //8 1, 1, 1, 1, 1, 1, 1 0000000
    {24}    //9 1, 1, 1, 0, 0, 1, 1 0011000
};


int i=0;
int j=0;
int k=0;
int l=0;
int m=0;
double ReadVal=1;
main()
{
    LJ_ERROR ljError; // LabJack error code
    LJ_HANDLE ljHandle = 0; // ID# assigned to the opened LabJack
    LoadLabJackUD(); // Load the LabJack DLL

// Open the first found LabJack U3
    ljError = OpenLabJack (LJ_dtU3, LJ_ctUSB, "1", 1, &ljHandle);
    ErrorHandler(ljError, __LINE__);

// Set all pin assignments to the factory default condition
    ljError = ePut(ljHandle, LJ_ioPIN_CONFIGURATION_RESET, 0, 0, 0);
    ErrorHandler(ljError, __LINE__);
    {
        while (ReadVal==1)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        while (m < 3300 && ReadVal==1)
        {
            i=k/330;
            j=l/33;

            if (j==10)
                l=0;


            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(1);
            k++;
            l++;
            m++;
        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }
        while (ReadVal==1)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(12);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(12);


        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        m=3300;
        k=3300;
        l=330;
        while (ReadVal==1 && m > 0)
        {

            i=k/330;
            j=l/33;

            if (l==0)
                l=330;


            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(1);
            k--;
            l--;
            m--;
        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        while (ReadVal==1)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(12);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(12);
        }

        ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
        ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
        return 0;
    }
}

engineering   Computer-Science   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

#include
#include
#include
#include
#include "LabJackUD.h"
#include "LJUD_DynamicLinking.h"


void ErrorHandler (LJ_ERROR ljError, long lngLineNumber)
{
    char err[255];
    if (ljError != LJE_NOERROR)
    {
        ErrorToString(ljError, err);
        printf("Error # %ld: %s\n", ljError, err);
        printf("Source line number = %ld\n", lngLineNumber);
        if(ljError > LJE_MIN_GROUP_ERROR)
        {
            getchar();
            exit(0); // Quit if serious error
        }
    }
}


const int numbers[10][1] =
{
    // 1: segment lit, 0: segment unlit
    //           a b c d e f g
    {64},   //0 1, 1, 1, 1, 1, 1, 0 1000000
    {121}, //1 0, 1, 1, 0, 0, 0, 0 1111001
    {36},   //2 1, 1, 0, 1, 1, 0, 1 0100100
    {48},   //3 1, 1, 1, 1, 0, 0, 1 0110000
    {25},   //4 0, 1, 1, 0, 0, 1, 1 0011001
    {18},   //5 1, 0, 1, 1, 0, 1, 1 0010010
    {2},    //6 1, 0, 1, 1, 1, 1, 1 0000010
    {120}, //7 1, 1, 1, 0, 0, 0, 0 1111000
    {0},    //8 1, 1, 1, 1, 1, 1, 1 0000000
    {24}    //9 1, 1, 1, 0, 0, 1, 1 0011000
};


int i=0;
int j=0;
int k=0;
int l=0;
int m=0;
double ReadVal=1;
main()
{
    LJ_ERROR ljError; // LabJack error code
    LJ_HANDLE ljHandle = 0; // ID# assigned to the opened LabJack
    LoadLabJackUD(); // Load the LabJack DLL

// Open the first found LabJack U3
    ljError = OpenLabJack (LJ_dtU3, LJ_ctUSB, "1", 1, &ljHandle);
    ErrorHandler(ljError, __LINE__);

// Set all pin assignments to the factory default condition
    ljError = ePut(ljHandle, LJ_ioPIN_CONFIGURATION_RESET, 0, 0, 0);
    ErrorHandler(ljError, __LINE__);
    {
        while (ReadVal==1)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        while (m < 3300 && ReadVal==1)
        {
            i=k/330;
            j=l/33;

            if (j==10)
                l=0;


            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(1);
            k++;
            l++;
            m++;
        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }
        while (ReadVal==1)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(12);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(12);


        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        m=3300;
        k=3300;
        l=330;
        while (ReadVal==1 && m > 0)
        {

            i=k/330;
            j=l/33;

            if (l==0)
                l=330;


            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(1);
            k--;
            l--;
            m--;
        }

        while (ReadVal==0)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
        }

        while (ReadVal==1)
        {
            ljError = eGet (ljHandle, LJ_ioGET_DIGITAL_BIT, 11, &ReadVal, 1);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[i][0], 7);
            Sleep(12);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 1, 0);
            ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_PORT, 0, numbers[j][0], 7);
            Sleep(12);
        }

        ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 7, 0, 0);
        ljError = ePut(ljHandle, LJ_ioPUT_DIGITAL_BIT, 8, 0, 0);
        return 0;
    }
}

0 0
Add a comment
Know the answer?
Add Answer to:
C Language: Write a diagnostic code for the following code. And please tell me what it...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • Help me to fix this code in C language. This code converts infix expressions to postfix and then evaluate the expression...

    Help me to fix this code in C language. This code converts infix expressions to postfix and then evaluate the expression. Right now, it works with single digits. I need to modify it so that it can evaluate expressions with also 2 digits, example: (60+82)%72. Additionally I need to display an error when the parenthesis don't match like (89+8(. I have muted some line that would print the postfix expression with a space like: 22 8 + when the input...

  • How would I be able to get a Merge Sort to run in this code? MY...

    How would I be able to get a Merge Sort to run in this code? MY CODE: #include <iostream> #include <fstream> #include <stdlib.h> #include <stdio.h> #include <time.h> using namespace std; class mergersorter { private:    float array[1000] ;    int n = 1000;    int i=0; public:    void fillArray()    {        for(i=1;i<=n;i++)        {            array[i-1]= ( rand() % ( 1000) )+1;        }    }    void arrayout ()    {   ...

  • should be in C language What is the output of the following segment of C code:...

    should be in C language What is the output of the following segment of C code: void CapsLetter (char* x, charl); int main() { char* text; text="This is some sample text."; CapsLetter(text, 'e'); printf("%s", text); return 0; سی void Caps Letter (char* x, char 1) { int i=0; while (x[i]!='\0') { if (x[i]==1) x[i]=1-32; } 1. This is som sample text. 2. THIS IS SOME SAMPLE TEXT. 3. This is some sample text. 4. this is some sample text. What...

  • Problem #2 (20pts) Please write the MIPS assembly code corresponding to the following C code segment....

    Problem #2 (20pts) Please write the MIPS assembly code corresponding to the following C code segment. Assume that f is assigned to register $s2 int Example int i, int j, int k) int f; f i+j-k-1: return fi

  • A) Rewrite the following pseudocode segment in C++ using the loop structures (for and while). Assume...

    A) Rewrite the following pseudocode segment in C++ using the loop structures (for and while). Assume that all the variables are declared and initialized. k-G+ 13)/27 loop: if k> 10 then goto out k=k+1 i=3"k-1 goto loop I out: B) Rewrite the following code segment in C++ using the multiple-selection (switch and if-else) statements. Assume that all the variables are declared and initialized. if((k 1) || (k 2))j-2 k-1 if ((k 3) || ( ks))j - 3.k+1 if (k 4)j-4k-1...

  • MSP430 Misc.peripherals. Consider the following C source code. What does the code segment from lines 9...

    MSP430 Misc.peripherals. Consider the following C source code. What does the code segment from lines 9 ~ 12 do? USCIA0 is configured in the UART mode. Consider the following C source code. 1. char gm1[]"MSP430"; 2. void UARTO_putdchar (char c) l 3. while (! (IFG2 & UCAOTXIFG)) UCAOTXBUF 5. while (! (IFG2 & UCAOTXIFG)) 6. UCAOTXBUF=''; 7 - 9. for (int i=0; i < 6; ¡++) { 10. ch gml [il; 11 UARTO putdchar (ch) 12. ) D. (4 points)...

  • 1. What is the output of the following code segment? int array[] = { 8, 6,...

    1. What is the output of the following code segment? int array[] = { 8, 6, 9, 7, 6, 4, 4, 5, 8, 10 }; System.out.println( "Index Value" ); for ( int i = 0; i < array.length; i++ ) System.out.printf( "%d %d\n", i, array[ i ] ); 2. What is the output of the following code segment? char sentence[] = {'H', 'o', 'w', ' ', 'a', 'r', 'e', ' ', 'y', 'o', 'u' }; String output = "The sentence...

  • Please use C programming to write a code segment as an answer. Instead of using an...

    Please use C programming to write a code segment as an answer. Instead of using an while-loop like the following code, please implement a code segment by using a do-while loop. What is the output of the code? #include <stdio.h> void main() {        int i = 0;        while (i < 5);        {               printf(“%d ”, ++i);        } }

  • This question is about the Arduino code. Please complete a 4-input circuit on your Arduino. Your...

    This question is about the Arduino code. Please complete a 4-input circuit on your Arduino. Your circuit, in addition to cycling through a binary representation of the numbers 0-15 using LEDs, should use the Serial port to output the decimal equivalent of each of the numbers in the format: Output: 15 ... if the value was 1111, and so forth. You will be asked to upload your code as well as a photo of your working breadboard circuit. Add comments...

  • Question 33 Consider the following code segment. String letters - ("A", "B", "C", "D"), ("E", "P",...

    Question 33 Consider the following code segment. String letters - ("A", "B", "C", "D"), ("E", "P", "G", "1"), ("I", "J", "K", "L"}}; for (int col = 1; col < letters[0].length; col++) for (int row - 1; row < letters.length; row++) System.out.print(letterstrow][col] + " "); System.out.println(); What is printed as a result of executing this code segment? AET PJ DHL G J K L PJ GK

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT