switch cases and one global variable for each case - c

Question

I am dealing with a issue with switch cases.
Explanation of the program:
main(argc,argv).
argv leads to cases in a switch statement. Depending on the input, the according case will be entered and the corresponding function will be executed. -> Output is always a struct, with different content. More than one input (i.e. main.c case1 case3) is allowed-> executed both cases.
my problem is dealing with the passing of these data's and save it in a global variable, in order to print the collection. Inside of a case, I am passing the local results to the global variable, but after the break statement of the case, the global starts with NULL again and doesn't contain the info's of the executed case.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "m.h"
output* print;
int main(int argc, char* argv[])
{
output_g* global; // global_struct
if(argc > 1)
{
global= create_global(argc-1); // allocation for global struct
for(int j = 0; j < argc; j++)
{
switch (atoi(argv[i]))
{
case 123:
{
output* print= 123();
if(print== NULL)
{
return 0;
}
global = fill_global(global ,print); // hands over the current struct to the global struct
delete(print); // free function
}
break;
case 456:
{
output* print= 456();
if(print== NULL)
{
return 0;
}
global = fill_global(global ,print); // hands over the current struct to the global struct
delete(print); // free function
}
break;
case 789:
{
lfnr++;
output_1* print_liste = 789();
if(print== NULL)
{
return 0;
}
global = fill_global(global ,print); // hands over the current struct to the global struct
delete(print); // free function
}
break;
default:
break;
}
print_global_struct(file,globale_liste);
delete_global(globale_liste);
}//End for-Schleife
}// End If
return 0;
}

Answer 1

a) If I understood you correctly, you don't understand the switch statement :)
A switch statement is similar to nested if statements.
so..
int x = 10;
switch (x)
case 1:
//does not happen, x is not 1
case 10:
//happens ...
after all that x is still 10, unless you changed it in the case statements explicitly. The cases just check to see IF x is a value, it does not SET a value. The same is true for any other variable in your code, the above code would not modify y either, unless you explicitly assign it inside a case it won't change.
b) it is best if you DO NOT declare locals in a case statement. They can become very wonky. The standard c++ rules work: variables declared inside {} pairs are scoped to inside that {} pair, so proper use of them will properly give the correct scope for each case. So it will work as expected if you apply braces. You should NOT declare a local in one case and use it in another, even if you can get it working (you can) it is error prone in that editing the code later can break things, the code can be confusing to read and work with, and its just generally a bad idea.
an example:
int main()
{
int x = 3;
switch(x)
{
case 1: int y;
case 2: y = 3;
case 3: y = 5;
cout << y << endl;
};
}
that compiled and ran for me, printing 5.
It worked fine -- I did not expect that, using g++ c17 options.
I still think it is a bad thing to do as far as reading and following the intent.
if you put {} around the int y statement, it does NOT compile anymore.
If you put breaks after the case 1 and case 2, it does NOT compile anymore.
so it is 'fragile' to being edited, at the very least, to do this.
c) run time won't lose anything. Ive had programs that ran for months on end. Being in a case has no effect on this either. The risk of losing variables is the 'ram' risk that all programs face... if a long running program is killed by power outage or malfunction etc you lose anything not saved to a file and have to start over. Long running programs should save their state periodically to protect against this.