Problems with switch...

AMMullan · 12-16-2003, 04:52 AM

K i'm newish to programming - C

Finally got round pointers and the one thing I thought I could do was switches, but i get the following when I try to compile my little app...

a2.c: In function `question':
a2.c:16: switch quantity not an integer

this is the code....

Code:

#include <stdio.h>
 
get_data(text)
char *text;
{
  printf("Enter the statement: ");
  scanf(" %s", text);
}
 
question(blah)
int *blah;
{
printf("Please select 1 - 5: ");
scanf(" %d", blah);
 
  switch(blah) {
    case '1':
      printf("Yes 1 would be good if you enjoy being single");
      break;
    case '2':
      printf("2 is good :)");
      break;
    case '3':
      printf("Haha 3 would be god, WAKE UP");
      break;
    case '4':
      printf("Haha now dreams may cost you");
      break;
    case '5':
      exit(1);
  }
}
 
int main() {
  int amount, i;
  char text_statement[100];
  int blah_blah;
 
  get_data(&text_statement);
 
  printf("How many times would you like to print the statement :: ");
  scanf(" %d", &amount);
  
  question(&blah_blah);
 
  printf("\nPrinting statement %d times\n", amount);
  for(i = 1;i <= amount;i++)
    printf("%s\n", text_statement);
}

Kumar · 12-16-2003, 05:25 AM

Hi,
try using
switch(*blah) in place of switch(blah) and it should work

and it should be case 1: instead of case '1': etc.

AMMullan · 12-16-2003, 05:33 AM

Yeah it worked perfectly, thanks

hmmm weird point tho, when i run it it doesn't come print the statement thats in the switch... weird?

Also, why would the switch statement need a pointer in front of the variable?

Tesl · 12-16-2003, 05:41 AM

the * itself isnt a pointer, you have declared

int * blah;

which means that the variable blah is the pointer. The pointer points to the parameter. When you called that function instead of passing an integer, you passed a reference (ie you passed the address of where an integer is stored).

This means that in the original switch statement, it was trying to switch a reference, or an address, which it cant do. When you use the pointer, it tries to switch the data that the pointer is pointing to. The pointer is pointing at that address, so it takes its (integer) value, which can then be switched

I really hope that made some sense, im in a big rush right now =/

by the way though, it didnt even compile at all for me when i tried it oO

AMMullan · 12-16-2003, 05:47 AM

hehe it compiled without any errors for me :-P

as for making sense, i got 99% of it, still reading and playing round with pointers so i'll get the rest eventually

thanks for the huge help guys