Thread: Making a switch error proof

I should've said from the beginning that the program would be a continuous loop, but I didn't think the loop would add such complications. My fault...

Code:

int main(int argc, char** argv) {

int exercises;
char str[50];

while (1) {
    printf( "1. Option 1\n"
            "2. Option 2\n"
            "3. Option 3\n"
            "4. Option 4\n"
            "5. Quit\n"
            "Choose a case: ");

    while( scanf( "%d", &exercises ) != 1 ) {
    char buf[ BUFSIZ ] = {0};
    fgets( buf, BUFSIZ, stdin );
}
            
  switch (exercises) {
    case 1: printf("1\n"); break;
    case 2: printf("2\n"); break;
    case 3: printf("3\n"); break;
    case 4: printf("4\n"); break;
    case 5: exit(EXIT_SUCCESS); break;
    default: printf("Invalid command.\n"); break;
     }
  }
}

I'm back to quzah's original suggestion. That bit of code does indeed read a number, and ignores everything else.

Since I'm using it, I should know exactly what it's doing. Can anyone explain it? Also, can I use sscanf() in this particular code to check if it's anything else other than a number?

First the char buf[busiz] should be outside your while loop... otherwise you're recreating it constantly. Also you need to define bufsiz 100 (or so) to hold the data. the part after the = sign initializes the buffers to all 0s.

What the fgets() does is bleed the input buffer dry so that the next scanf() call sees an empty input queue.

Also, a minor issue, you don't always need all the braces you use. It's mostly harmless to have them in there but there are some infrequent cases where it might matter, so best learn when to and when not to use them.

Code:

#include <stdio.h>
#define BUFSIZ 100

int main(int argc, char** argv) {

   int exercises;
   char str[50] = {0};
   char buf[ BUFSIZ ] = {0};

while (1) {
    printf( "1. Option 1\n"
            "2. Option 2\n"
            "3. Option 3\n"
            "4. Option 4\n"
            "5. Quit\n"
            "Choose a case: ");

    while( scanf( "%d", &exercises ) != 1 ) 
       fgets( buf, BUFSIZ, stdin );

  switch (exercises) {
    case 1: printf("1\n"); break;
    case 2: printf("2\n"); break;
    case 3: printf("3\n"); break;
    case 4: printf("4\n"); break;
    case 5: exit(EXIT_SUCCESS); break;
    default: printf("Invalid command.\n"); break;
     }
  }
}

There are, of course, other solutions you can use here instead of scanf()... getchar() and fgetc() come to mind.

One of the truisms in programming is that input is always harder than output... you never know what some silly operator is going to type into your code. I've even seen cases where operators have deliberately fed in wrong or corrupt data so they could make a service call on their software... and get the afternoon off.

Code:

while( scanf( "%d", &exercises ) != 1 )

What is this checking?

So what is it returning if I enter, say, 2?

What does the 1 represent, and why would I have a problem if it's returning something other than 1, if the while loop is testing for everything other than 1?

Originally Posted by never_lose

So what is it returning if I enter, say, 2?

What does the 1 represent, and why would I have a problem if it's returning something other than 1, if the while loop is testing for everything other than 1?

If you enter 2, the 2 is stored in the pointer you pass in and the function returns 1, because that's the number of things it stored. Since 1 == 1, you skip the loop and move on with your life.

Originally Posted by grumpy

I can't say I've seen people doing it for time off. I have, however, seen cases where a bored operator doing a routine job does starts playing with different inputs in order to see what happens.

I visited a conference stand a few years ago, and a salesman quite proudly told me that the GUI was well engineered and unbreakable, so I could do what I liked. I closed my eyes, placed my hand somewhere on the keyboard, and pushed down (hence depressing several keys at once). The screens went black, and then the software crashed leaving the underlying X-windows on screen. I then exited stage left.

Apparently it was about three hours before they got an engineer in to get the software going again.

No joke... I have a paint roller (clean and dry of course) that I run across the keyboard when writing console programs... just to test this very thing!

Originally Posted by quzah

No it doesn't. Even if we assume it does, all that means is that it would nicely clear your buffer every time the loop happened.No, it's a standard macro.
Quzah.

As I have just now discovered ... Hey, I don't have the headers memorized, you know!

Hmm, I've got a pretty good solution now, but not completely error proof.

Look what happens if I enter "1 3".

Choose a case: 1 3
1
1. Option 1
2. Option 2
3. Option 3
4. Option 4
5. Quit
Choose a case: 3
1. Option 1
2. Option 2
3. Option 3
4. Option 4
5. Quit

Code:

#include <stdio.h>
#include <stdlib.h>
//#define BUFSIZ 100

int main(int argc, char** argv) {

   int exercises;
   char str[50] = {0};
   char buf[BUFSIZ] = {0};

while (1) {
    beg:
    printf( "1. Option 1\n"
            "2. Option 2\n"
            "3. Option 3\n"
            "4. Option 4\n"
            "5. Quit\n"
            "Choose a case: ");

    while ( scanf( "%d", &exercises ) != 1 ) {
       fgets( buf, BUFSIZ, stdin );
    printf("Enter a number!\n");
    goto beg;
}

  switch (exercises) {
    case 1: printf("1\n"); break;
    case 2: printf("2\n"); break;
    case 3: printf("3\n"); break;
    case 4: printf("4\n"); break;
    case 5: exit(EXIT_SUCCESS); break;
    default: printf("Invalid command.\n"); break;
     }

  }
}

I commented off #define BUFSIZ since I don't need it, right?

For understanding purposes, if I entered a letter, would it return 0, since "scanf( "%d", &exercises )" can't store that letter as a number?

Originally Posted by never_lose

Hmm, I've got a pretty good solution now, but not completely error proof.

Look what happens if I enter "1 3".

If that's bothersome to you, you can run through the bleed-the-buffer-dry after every input, not just unsuccessful input.

Originally Posted by never_lose

For understanding purposes, if I entered a letter, would it return 0, since "scanf( "%d", &exercises )" can't store that letter as a number?

scanf always returns the number of inputs it was able to process (or EOF if there was error in even trying to read anything at all). In this case, since it processed zero inputs, and your keyboard actually exists, it will return 0.

Putting a "fgets( buf, BUFSIZ, stdin );" in every case solved that, but that's a lot of duplicate code. I'm thinking there is a better way to do it?

Now that you have a pile of code that does one particular thing ("read in a number"), that would be an excellent thing to turn into a function.