Login or Register to Ask a Question and Join Our Community


Programming

While loop problem in C

Login to Discuss or Reply to this Discussion in Our Community

 
Thread Tools Search this Thread
Top Forums Programming While loop problem in C
# 1  
Old 11-04-2012
While loop problem in C
Hello friends,
Please have a look at this code, and then i will tell you the problem.

Code:
 
int main()
{
    int i = 0;
    int cont = 1;
 
    while( cont == 1 ) 
    {
             printf( "Enter i\n" );
             scanf( "%d", &i );
 
             switch( i )
             {
                    case 1:
                              printf( "1\n" );
                              cont = 0;
                              break;
                    case 2:
                              printf( "2\n" );
                              cont = 0;
                              break;
                    default:
                              printf( "Invalid value" );
             }
    }
 
    return 0;
}

Here is the problem. The condition for while loop is true initially, so it's body is executed. But when the default case for switch executes, control returns to while, and while the condition is true, the body of while does not execute, and the program does not terminate. What could be the problem?

Thanks in advance!
# 2  
Old 11-04-2012
Firstly I personally not understanding the way u have stated the problem and what doubt you have. The prog works perfectly, when default statement executed the 'cont' value doesn't get changed, remains 1 so control gets returned to while. its fine. Whats wrong in that?
These 2 Users Gave Thanks to amisubha For This Post:
gabam Raj 89
# 3  
Old 11-04-2012
Add the break; clause in the default section of the switch ?
# 4  
Old 11-05-2012
You are using scanf, which is liable to have problems. If it sees any input it doesn't like -- any non-numbers here -- it will not skip them. They will remain in the buffer, with scanf returning an error code every loop, and your loop will become an infinite one.

Never use scanf. Use sscanf and fgets. It won't choke on input, bad lines will just be skipped.

Code:
char buf[512];
fgets(buf, 512, stdin);
sscanf(buf, "%d," &intvalue);

# 5  
Old 11-23-2012
hmm, when input is not 1 or 2, it seems like the while loop is true.
It looks right to me...
# 6  
Old 11-28-2012
Quote:
Originally Posted by gabam

Code:
 

    int cont = 1;
 
    while( cont == 1 )   
    {
             printf( "Enter i\n" );
             scanf( "%d", &i );
 
             switch( i )
             {
                    case 1:
                              printf( "1\n" );  
                              cont = 0;
                              break;
                    case 2:
                              printf( "2\n" );
                              cont = 0;
                              break;
                    default:
                              printf( "Invalid value" );
             }
    }

Your program is working perfectly fine.I am not able to understand your logic clearly. What is that you are trying to do. You have normally written the while loop in such a way that it will be exited whenever the value of cont is other than 1. So at first the while loop satisfies and the user provides the input for "i".
But when you provide 1 or 2 for the switch case, the value const will be replaced with zero and then the execution goes to while. This will act as an infinite loop till the value of i tends to be other than 1 r 2 .
but if the value of i is other than 1 or 2, cont will be same and the while loop fails the second time.

---------- Post updated at 03:17 PM ---------- Previous update was at 03:15 PM ----------

Quote:
Originally Posted by gabam

Your code is replaced into this. Well this works better.

Code:
 
int main()
{
    int i = 0;
    int cont = 1;
 
    while( cont == 1 ) 
    {
             printf( "Enter i\n" );
             scanf( "%d", &i );
 
             switch( i )
             {
                    case 1:
                              printf( "1\n" );
                              break;
                    case 2:
                              printf( "2\n" );
                              break;
                    default:
                              printf( "Invalid value" );
                              cont = 0 ;
             }
    }
 
    return 0;
}

Maybe this could help you...
Login or Register to Ask a Question

Previous Thread | Next Thread

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Problem with loop

Hi all, a problem with a loop. Imagine it starts with var1="$(cat txtif.out )"while do echo file1 echo file2 echo y > txtif.out if then break fi done exit 0 the problem is that if the file is changing during the loop seems to continue as it reads "x" state,... (5 Replies)
Discussion started by: Board27
5 Replies

2. Shell Programming and Scripting

Problem with While Do loop

I am trying to do a while do loop that asks for a password and returns if the password entered is incorrect. I have that part working however i want to modify it so that if a similar word is entered it will read "that is close but incorrect" and i cannot seem to get it working, being new in the... (6 Replies)
Discussion started by: lm5522
6 Replies

3. Shell Programming and Scripting

Problem with loop within loop

Hi, I work on Ab-initio ETL tool which is based on Unix. I made a small script which has two loop's one with in another. All the functionality is working for the first line of outer loop but when it comes to other lines of outer loop it is throwing error as command not found. Below is the... (4 Replies)
Discussion started by: Ravindra Swan
4 Replies

4. Shell Programming and Scripting

Problem with While loop.

Hi Script Gurus, I am facing issue with while loop in bash. The while loop is running as end less loop even after given criteria does not meet. STATE_BEFORE_SPLIT () { for clone in $Clonegroups; do state=$( $Navicmd -listclone -name $clone -cloneid $Cloneid |awk... (3 Replies)
Discussion started by: RobP
3 Replies

5. UNIX for Dummies Questions & Answers

While loop problem

I have a while loop with -f and -o option.Can any one please tell me what that stands for?The Sample code is as follows:- while do ## some processing done (10 Replies)
Discussion started by: Param0073
10 Replies

6. Shell Programming and Scripting

problem with while loop

hi I had created a while loop in a script file called whiletest.sh as follows as follows: count=0 max=10 while do echo $count echo count=$(count+1)" " done echo "value of count:$count" then i run the script with the command sh whilestest.sh but its giving me an error... (5 Replies)
Discussion started by: angel12345
5 Replies

7. Shell Programming and Scripting

Problem in While loop

Hi Guys, This is my code.. I am getting an error in the inside while statement when comparing.. I am not able to figure it out.. Pls help me... while read value fatherid son_id top_id r_type do inside_value=1; echo $inside_value; echo $r_type; while do echo... (6 Replies)
Discussion started by: mac4rfree
6 Replies

8. UNIX for Advanced & Expert Users

loop problem

Hi guys my while do loop is not working properly; As soon as the load_date and run_date is same it should stop can somebody tell me where I am having the problem? In an oracle table I have LOAD_DT=5/1/2009 DATE datatype RUN_DT=5/5/2009 DATE datatype Now in a script with a spool file I get the... (15 Replies)
Discussion started by: henrysmith
15 Replies

9. Shell Programming and Scripting

for loop problem

Hi, I have a directory called logs in which i have the log files. i have to touch the file before deleting it. i am doing like this filestodelete="*.log* *log*" for files in $filestodelete do touch $files $files.$(date +%a) rm -f $filestodelete done touch is not working... (5 Replies)
Discussion started by: namishtiwari
5 Replies

10. Programming

problem with while loop

hi all, i have written the following code: while(proceed !='Y' && proceed!='N' && proceed!='y' && proceed!='n') { printf("\nPress \n\t 'Y' or 'y' to continue \n\t 'N' or 'n' to cancel:"); scanf("%c",&proceed); } the output i am gettin is: Press 'Y' to continue ... (1 Reply)
Discussion started by: mridula
1 Replies
Login or Register to Ask a Question