How do I run this without the loop printing the "error" message

Question

So I am very new to coding (literally started learning c a few days ago) and I decided to just play around and see if I can apply what I learned so far. I created an "employee search" program that prompts the user for a name and it will check to see if the employee exists. I ran into an issue within the loop; if I were to type "Chris" into the terminal and click enter, it'll say something like: "Employee not found." "Chris found." "Employee not found."How do I go about making the program confirm the name is within the "database" without it repeating the "error" message. Sorry for the newbie question. Again, I'm very new to this.

#include <cs50.h>
#include <stdio.h>
#include <string.h>

int main(void)
{
    // declare array
    string employee[] = {"Damien", "Chris", "Emma"};

    // print intro message and prompt user for name
    printf("Welcome to employee search\n");
    printf("Please input an employee name: ");
    string name = get_string();

    // here is where I run into the issue where it'll repeat "employee not found"
    for(int i = 0; i < 3; i++)
    {
        if(strcmp(name, employee[i])==0)
        {
            printf("%s found\n", name);
        }
        else
        {
            printf("Employee not found\n");

        }
    }
}

What is the line `found."` doing? Is this C i.e. what is `string`? — Ed Heal, Nov 25 '17 at 07:25
By setting a flag within the loop and reporting afterwards. Instead of reporting on each iteration. — Weather Vane, Nov 25 '17 at 07:26
... or even better, you could perhaps easily find the *answer* on cs50.stackexchange.com without asking — Antti Haapala -- Слава Україні, Nov 25 '17 at 07:26
It all depends on whether the unknown `get_string();` function includes the *line termination* character. This is also a good example of whether [Is it a good idea to **typedef** pointers?](http://stackoverflow.com/questions/750178/is-it-a-good-idea-to-typedef-pointers). (which apparently is occurring in `cs50.h`) — David C. Rankin, Nov 25 '17 at 07:30
@DavidC.Rankin "all depends": no - `get_string` is called *once* and he is reporting in each iteration. — Weather Vane, Nov 25 '17 at 07:32
What I meant is you will never match it the line endings are included... — David C. Rankin, Nov 25 '17 at 07:33
That's true, such as `fgets` but his string *does* match. That is not the issue. — Weather Vane, Nov 25 '17 at 07:33
That answers the "Are they included?" question then. I must be getting punchy... — David C. Rankin, Nov 25 '17 at 07:34
@DavidC.Rankin - My curiosity has got the better of me. What does the "c" stand for? — Ed Heal, Nov 25 '17 at 07:38
Hmm... You have me at a loss Ed? If it is the `c` in `"punchy"` it means loopy, tired, bewildered and without it, it would be a awkward spelling of the metaphor getting smaller `:)` Oh lightbulb on ... Charles... (I told you I was getting punchy...) — David C. Rankin, Nov 25 '17 at 07:42
I apologize for bringing this question on here. I just hopped on cs50 earlier this week to learn more before I head off into a CS program at university. I didn't know cs50 had their own stackexchange forum but I'll move my questions there. @AnttiHaapala — ApacheR12, Nov 25 '17 at 07:43

score 2 · Accepted Answer · answered Nov 25 '17 at 07:28

Avoid printing inside the loop. Instead use a flag to save the status. Like:

int flag = 0;  // Initialize flag to 0 (i.e. assume the name isn't found)

for(int i = 0; i < 3; i++)
{
    if(strcmp(name, employee[i])==0)
    {
        flag = 1;  // Set flag to 1 to remember that we had a match

        break;     // Stop the loop using break. We don't need to check the rest
                   // as we have found a hit
    }
}

if (flag)
{
    printf("%s found\n", name);
}
else
{
    printf("Employee not found\n");
}

Zahid Khan · Answer 2 · 2017-11-25T09:30:44.080

1

#include <cs50.h>
#include <stdio.h>
#include <string.h>

int main(void)
{
   string employee[] = {"Damien", "Chris", "Emma"};
   int i = 0;

// print intro message and prompt user for name
 printf("Welcome to employee search\n");
 printf("Please input an employee name: ");
 string name = get_string();       //Declaration and initialisation

for(i = 0; i < 3; i++)
{
    if(strcmp(name, employee[i])==0)
    {
        printf("%s found\n", name);
        break;   // if any of the employee is found it exit the loop immediately with the value of i<3 
    }

}
if (i ==3 ) //means the loop has reached end without finding any of employee.
        printf("Employee not found\n");     
}

edited Nov 25 '17 at 09:30

answered Nov 25 '17 at 07:30

Zahid Khan

2,130
2
18
31

Perhaps format the code properly and add an explaination – Ed Heal Nov 25 '17 at 07:39
Added explanation, hope it help yew – Zahid Khan Nov 25 '17 at 07:56
1

`i` is out of scope outside the `for` loop. – David Ranieri Nov 25 '17 at 07:57
It works fine that way too in codeblock. Still made modification. – Zahid Khan Nov 25 '17 at 09:32

score -2 · Answer 3 · answered Nov 25 '17 at 07:37

-2

 // yes dear it a bit easy  
 // first get the input in any variable 
  string name="";
 // And  then inside a loop chek every index of array to the string which you get from the user  

cout<<" Enter name ";
 cin>>name;

 for(int i=0; i<=c.length;i++)
{  if(c[i]==name) 
    {  
     cout<<"found";
   }
else{  cout<<"not found ";
   }

}

// c just like c={"first name","secound_name","blablala"}

answered Nov 25 '17 at 07:37

Harendra Singh Rawat

9
3

Perhaps format the code properly and add an explaination – Ed Heal Nov 25 '17 at 07:39
1

Question is tagged c not c++ – David Ranieri Nov 25 '17 at 07:41
This code does not provide an answer to the question, it faced the same issues if it'd work. – alk Nov 25 '17 at 10:34

How do I run this without the loop printing the "error" message

3 Answers3