Understand backspace (\\b) behaviour in C

Question

This program copy its input to its output, replacing TAB(\t) by \t backspace(\b) by \b. But here in my code I am unable to read input character when I enter backspace its not replacing as a tab works .

Compiling with GCC in Linux:

#include<stdio.h>
int main(void)
{
    int c=0;
    while((c=getchar())!=EOF){
     if(c=='\t'){
      printf("\\t");
     if(c=='\b')
      printf("\\b");
    }
    else
     putchar(c); 
}
return 0;
}

Suppose if I type vinay (tab) hunachyal

Output:vinay\thunachyal 

If I type vinay(and 1 backspace)

Output:vina

So my query is why vina\b is not printing in this case?
Is it possible to detect \b and print \b? if not whats the reason

Note: I need at run time input backspace not providing separate file having \b

Answer 1

The backspace is consumed by the shell interpreter, so your program will never see it, also your code is (slightly) broken, due to misplaced braces, not helped by the poor indentation.

Here is a corrected version:

#include<stdio.h>
int main(void)
{
    int c=0;
    while((c=getchar())!=EOF){
        if(c=='\t')
            printf("\\t");
        else if(c=='\b')
            printf("\\b");
        else
            putchar(c);
    }
    putchar('\n');
    return 0;
}

which works as expected:

$ echo 'vinay\thunachyal\b' | ./escape
vinay\thunachyal\b

Answer 2

If I haven't misinterpreted the question, you may use 'Ctrl-H' to send a backspace. Using trojanfoe's corrected code, when you type:

vinay^H

It will print:

vinay\b

^H means 'Ctrl-H', it's ASCII character #8, which is backspace.