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    Telecomunication program - pb with how to verifie that a nbr is included in an array-


    I've been asked to write a program for which :

    You are asked to write a C program for telecommunication company, the program 1- asks the user to enter his number.
    2- displays which operator he is using.
    3- asks him if: -he wants to buy credit ( in this case asks for the amount of money).
    - he wants to talk to an agent ( and keeps talking until he presses 1 to hung out)
    - he wants to send a default message for a friend who uses the same operator ( he should enter the number of the friend and the program verifies if it is the same operator, and displays a choice of four default messages [you are free to choose default messages] )

    -he wants to exit.
    The user should be able to make diffrent operations without closing the program, whenever an operation is finished , the programs goes back to the main menu.The program only exits when the user chooses “exit” from the menu.



    my pbs were : 1-I don't know how to ( make the compiler verifie if it is the same operator )

    2- while I am using the ( do while ) it keeps reapiting the " enter ur number " and when I switched it below it keeps reapiting only the 1st condition




    this is my program if anyone could help me :


    Code:
    #include <stdio.h>
    #include <stdlib.h>
    int number,n,friendnumber,amountofmoney;
    int operatorr;
     
    int main()
    {
    do
        {
        printf("Enter your number please :\n");
        scanf("%d",&number);
        printf("You are now using the operator: 123\n");
     
        printf("press 2 to choose how to buy and 3 to talk to an agent and 1 to send a default msg and 0 if you want to exit ");
        scanf("%d",&n);
     
     
            if(n==2)
        {
     
              printf("\nyou will buy credit, enter your amount of money :");
              scanf("%d",&amountofmoney);
     
        }
        else if  (n==3)
        {
            printf("\nyou are now talking to an agent press 1 to hung out");
            scanf("%d",&n);
        }
        else if (n==1)
        {
            printf("\nif you want to send a default message to a friend enter his number : ");
            scanf("%d",&friendnumber);
            if (friendnumber[&number])
            {
                printf("Hi, how are you ?");
                printf("\nI need you in emergency");
                printf("\nlet's meet tonight");
                printf("\nI will be there ASAP");
            }
           else
           printf("\nyou're user is not using the same operator as you ");
        }
        }
        while (n!=0);
     
    return 0;
    }
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    here's something I did with some additions:


    Code:
    #include <stdio.h>
    #include <stdlib.h>
    int number,n,friendnumber,amountofmoney;
    int operatorr,foperatorr;
    long a,b,c,d,e,f;
    int x;
    
    char opa[]="operator 1";//number beginning with 100
    char opb[]="operator 2";//number beginning with 200
    
    int main()
    {   printf("Enter your 7 digit number please :\n");
        scanf("%d",&number);
        a=(number/1000000);
        b=(number/100000)-(a*10);
        c=(number/10000)-(a*100)-(b*10);
        {{if(a==1)
        {
            if(b==0)
            {
                if(c==0)
                    {operatorr=1;
                    printf("you are using operator %d\n\n",operatorr);}
            }
        }}
    
        {
        if(a==2)
    
        {
            if(b==0)
            {
                if(c==0)
                    {operatorr=2;
                    printf("you are using operator %d\n\n",operatorr);}
            }
        }
        }}
    {do
        {
    
        printf("press 2 to choose how to buy and 3 to talk to an agent and 1 to send a default msg and 0 if you want to exit ");
        scanf("%d",&n);
    
    
            if(n==2)
        {
    
              printf("\nyou will buy credit, enter your amount of money :");
              scanf("%d",&amountofmoney);
    
        }
        else if  (n==3)
        {
            printf("\nyou are now talking to an agent press 1 to hung out");
            scanf("%d",&n);
        }
        else if (n==1)
        {
            printf("\nif you want to send a default message to a friend enter his number : ");
            scanf("%d",&friendnumber);
            d=(friendnumber/1000000);
            e=(friendnumber/100000)-(d*10);
            f=(friendnumber/10000)-(d*100)-(e*10);
        {{if(d==1)
        {if(e==0)
        {if(f==0)
        foperatorr=1;}}}
        {if(d==2)
        {if(e==0)
        {if(f==0)
        foperatorr=2;}}}}
    
            {if(foperatorr==operatorr)
            {printf("you and you're friend are on the same operator.");
            printf("\nPlease choose from the following default messages.\n\n");
            printf("1. Hi, how are you ?");
            printf("\n2. I need you in emergency");
            printf("\n3. let's meet tonight");
            printf("\n4. I will be there ASAP");
            scanf("%d",&x);
            printf("\nmessage number %d sent to your friend.\n",x);}
            else
           printf("\nyou're friend is not using the same operator as you\n ");
        }}}
        while (n!=0);
        }
    }
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    Originally Posted by salem
    that is also un-answered...and probably posted by the same user ( just 4 hrs ago )

    so.......?
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    So....

    Asking in more than one place

    Asking the same question concurrently

    > and probably posted by the same user ( just 4 hrs ago )
    It's a FREE service, open 24 hours a day, 365 days a year.
    Normal etiquette would be that posters would not abuse that by posting the same thing over and over.

    Imagine if they'd got the same answer almost immediately - someone would have wasted their time answering the question.

    If no one replies in a reasonable time frame (at least 24 hours), then finding another forum is permissible, providing that the questions are then cross-linked for reference by the original poster.

    If you want answers in 4 hours, then pay someone who offers a quality of service guarantee.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper
  10. #6
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    Originally Posted by salem
    So....

    Asking in more than one place

    Asking the same question concurrently

    > and probably posted by the same user ( just 4 hrs ago )
    It's a FREE service, open 24 hours a day, 365 days a year.
    Normal etiquette would be that posters would not abuse that by posting the same thing over and over.

    Imagine if they'd got the same answer almost immediately - someone would have wasted their time answering the question.

    If no one replies in a reasonable time frame (at least 24 hours), then finding another forum is permissible, providing that the questions are then cross-linked for reference by the original poster.

    If you want answers in 4 hours, then pay someone who offers a quality of service guarantee.
    ok i get it...valid point.

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