Redundant question please delete

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1.20/5 (3 votes)

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/* program that uses the switch statement
*/
#include <stdio.h>

int main()
{
char oper;

printf("enter a mathematical operator \n");

oper=getchar();

//use the switch statement
switch(oper)
{
case'+':
{
printf("you entered a plus \n");
break;
}

case'-':
{
printf("you entered a minus \n");
break;
}

case'*':
{
printf("you entered a multiply \n");
break;

}

case'/':
{
printf("you entered a divide \n");
break;
}

default:
{
printf("invalid operator entered \n");
break;
} //end default

} //end switch
return 0;

} //end main

What I have tried:

Dont even know where to start. I'm just so confused.

Posted 12-Nov-16 5:07am

Tiarnan Mc Gurk

Updated 13-Nov-16 11:05am

v2

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OriginalGriff · Answer 1 · 2016-11-12T05:17:00

The error you have noticed is simple: You forgot the semicolon at the end of this line:

C++

char goAgain='y'

Adding it will fix the compilation problem:

C++

char goAgain='y';

But...then you have more problems.
1) You don't read the choice from the user.
2) You break out of the switch regardless of the user selection of "y" or "n".
3) Your condition doesn't check the user PIN is 1234, it assigns it. "=" is assignment, "==" is comparison.
4) You should be using your "goAgain" variable in the loop termination condition.

Suvendu Shekhar Giri · Answer 2 · 2016-11-12T05:27:00

Solution 2

You have used the variable choice before initializing the value and you have also forgotten to write the default case.

The possible solution could be to accept user input just before the switch statement.

Something like-

C++

//--your other previous stuffs
printf("4. Exit \n");
scanf("%d",choice); //accept choice from user

    switch(choice)
    {
       //---rest of the logic

Hope, it helps :)

Posted 12-Nov-16 5:27am

Suvendu Shekhar Giri

Comments

Suvendu Shekhar Giri 12-Nov-16 11:30am

Please consider the first solution and if you still get any issue, let us know.
Please ignore my solution for now.