Fixing C switch statement function overflow? - c

Question

int main (void)
{
*/ function prototypes */
print_case();
do_something1();
do_something2();
do_something3();
do_something4();
exit_program();
program_invalid();
}
void print_case (void)
{
int i;
printf("\n"
"1. Do Something 1\n"
"2. Do Something 2\n"
"3. Do Something 3\n"
"4. Do Something 4\n"
"5. Exit the program\n"
"Enter choice (number between 1-5)>\n");
scanf("%d", &i);
switch(i)
{
case 1:
do_something1();
break;
case 2:
do_something2();
break;
case 3:
do_something3();
break;
case 4:
do_something4();
break;
case 5:
exit_program();
break;
default:
program_invalid();
break;
}
return;
}
something_t do_something1(void)
{
something_t something;
printf("Something 1\n");
return something;
}
void do_something2(something_t something)
{
printf("Something 2\n");
}
void do_something3()
{
printf("Something 3\n");
}
void do_something4()
{
printf("Something 4\n");
}
void exit_program (void)
{
exit(0);
}
void program_invalid (void)
{
printf("Not valid choice");
}
So basically when I compile it and execute the code and select the various cases, it will execute multiple functions at once and prints out multiple statements at once. Let's say I choose case 1 the output it prints Something 1 but when I choose case 2 it prints
Something 1
Something 2
and when I choose case 3 it prints
Something 1
Something 2
Something 3
So how would I fix my code to get out of the loop? I thought break statements would only let it execute one function at a time. Yes the something_t references to my typedef structures that I didn't include in my code.

Answer 1

print_case() has the switch. It does its thing then returns. What you THINK are function prototypes in main() are actually just calls. So it calls them. And so you see all of the functions executing. C has a habit of shrugging and making that work, because traditionally it is very tolerant. Move your 'prototypes' out to before main() and preferably put a proper signature on them all.
Your do_something2 has an arg, but you are not declaring it in the (non-working) fake prototype - that is, it will be incorrect once you move it out to before main().
Also, since you have declared do_something2() to take an arg, you'd better pass one!

Answer 2

why are you putting something_t as input into your function. The code your posting also will not compile.
you also have a gap in the name and are missing a function type for function something_t do_something1(void).

Answer 3

Here is the clean version of your code , I think this might help you remember some stuff.
#include <stdio.h>
#include <stdlib.h> // for exit.
#define True 1 // Symbolic constants.
/*
* This is a multi-line comment.
* -----------------------------
* These is how a function prototype
* should be.
* You can leave the parameter names
* empty because in function prototypes
* the parameter names are dumy but not
* the types.
*/
typedef int something_t; // You can have any structure here.
// just for an example.
void print_case(void);
something_t do_something1(void);
void do_something2(something_t);
void do_something3(void);
void do_something4(void);
void exit_program (void);
void program_invalid (void);
// ---- This is a single line comment.
int main()
{
while(True) {
print_case();
}
return 0;
}
void print_case (void)
{
int i;
printf("\n"
"1. Do Something 1\n"
"2. Do Something 2\n"
"3. Do Something 3\n"
"4. Do Something 4\n"
"5. Exit the program\n"
"Enter choice (number between 1-5)>\n");
scanf("%d", &i);
switch(i) {
case 1:
do_something1();
break;
case 2:
do_something2(True); // must pass your struct.
break;
case 3:
do_something3();
break;
case 4:
do_something4();
break;
case 5:
exit_program();
break;
default:
program_invalid();
break;
}
return;
}
something_t do_something1(void)
{
something_t something;
printf("Something 1\n");
return something;
}
void do_something2(something_t something)
{
printf("Something 2\n");
}
void do_something3(void)
{
printf("Something 3\n");
}
void do_something4(void)
{
printf("Something 4\n");
}
void exit_program (void)
{
exit(0);
}
void program_invalid (void)
{
printf("Not valid choice");
}