Why does printf(“%d\n”, printf(“%d\b”, a)) work this way?

Question

This is my C code, compiled with gcc.

#include<stdio.h>

int main() 
{ 
    int a=1; 
    switch(a)
    {
       int x=10;
       case 1:
           printf("%d\n",printf("%d\b",x));
           break;
       default:
           printf("%d\n",printf("%d\b",x));
    }
    return 0;
}

printf() is supposed to return the number of elements it printed successfully. printf("%d\\b", x) should have printed 10 by itself(since the \\b takes the printing pointer one step behind (to the digit 0 in 10) and there is nothing to print after that. So it should have just printed 10. That is 2 characters. Now the outer printf would display 2. The output should have been 102. The output I actually see is 2.

And in case of nested printf s is the printing pointer position remembered? I mean, if there is a \\b in the inside printf , it would take the printing pointer one step behind. And when the control now goes to the outer printf , is that changed position remembered? Will it overwrite over that last character?

Answer 1

printf("%d\b",x)

prints the characters '1' , '0' (because x==10) and \\b . The \\b is a backspace character; if you print to a terminal, it will print 10 and then move the cursor back one column.

A call to printf returns the number of characters it printed; in this case, the result is 3 (yes, '\\b' counts as a character).

printf("%d\n",printf("%d\b",x));

The inner printf call works as I explained above, and returns 3. The outer printf call prints "3\\n" .

So the entire statement will print:

10\b3\n

The '\\b' causes the 3 to be replace the 0 on the screen, so the final displayed result (when I run the program on my system) is:

13

If I pipe the output through cat -v , I get:

10^H3

where ^H represents the backspace character.

EDIT :

The question was just edited, and the modified program's behavior is quite different. The switch statement causes control to jump past the declaration int x = 10; , but into the scope in which x is declared. As a result, x is uninitialized when printf is called. This causes undefined behavior, and most likely garbage output (I just got -1217572876^H12 ). If x happens to be 0 , I suppose you'd get 0^H2 , which would look like 2 .

Whatever you're trying to do, please find a better way to do it.