How array store in stack ..?

Hello all,

I stuck to a problem

Statement: i define two function one returning a string and other a single char.

the code is work fine it successfully return string "martin" and a char 'a'....

Code:

char *func(void)
{
     static char Text2[10]="martin";
     return (Text2);        
}
   
char* funchar(void)
 {
        char ch;
        
        ch = 'a';
        return &ch;
 }

main()
   {
      char *Text1;
      char *Text2;
      char* c;
      Text1 = func();
     
      printf("\ntext1:%s",Text1);
          
      Text2 = "outside";
      printf("\ntext2:%s",Text2);
      
      c = funchar();
        printf("\nc:%c",*c);

                      
      getch();
   }

but if i remove the word static from func(),like,

Code:

char *func(void)
{
     char Text2[10]="martin";
     return (Text2);        
}

than this function return a garbage but as in funchar() i didn't declare char ch as static still it return 'a'.

Why this is happen..?

Do array whether it is a char array or and other(int ,double) , only its location is stored in stack means actually the string "martin" stored somewhere else but its reference will stored in stack.

Or the whole string "martin" will stored in stack.?

Thanks,
Gunjan

Question 7.5a
string literal like "helo"are stored not in stack.
So it is valid to say return "hello".
But don't get confused with array initialization.a[]="hello". Which is same as a[]= { 'h','e','l','l','o',\0'}

funchar() is broken in the same way that func() is broken, when you remove the static.

In both cases, you've got a pointer to a local variable that has gone out of scope.

The memory is still there (most likely), and it may indeed print the value you expected, but the code is surely broken.

Thanks all for their quick inputs..