Question

int main()
   {       char *arr="\\0\1\8234\0"; int i=0;
           while(arr[i])
           {        switch(arr[i])
                    {
                            case '0': printf("is no"); break;
                            case '00': printf("is debugging\n"); break;
                            case 0: printf("It is Avishkar\n"); break;
                            case '\\': printf("This "); break;
                            case '\1': printf("t s"); break;
                            case '8': printf("o s"); break;
                            case '2': printf("imp"); break;
                            case '3': printf("le as"); break;
                            case 2: case 3: case 4:
                            case 8: printf("This "); break;
                            default: printf(" it seems\n"); break;}
                    i++; } }

please explain the o/p ?
i am not able to get it..

Answer 1

Breaking down the string initialization we have:

\\ == literal \ character
\1 == octal representation of integer "1"
\8 == technically invalid standard C, microsoft compilers will treat an
unsupported (i.e. non-octal, non-escape, non-hex-specifier) character after
a slash as if the slash were not present, so this is ascii '8'
2 == ascii '2'
3 == ascii '3'
4 == ascii '4'
\0 == literal zero

arr[ 0 ] = '\\';   // literal backslash
arr[ 1 ] = '0';    // ascii '0'
arr[ 2 ] = 1;      // integer 1
arr[ 3 ] = '8';      // using microsoft (non-portable) syntax
arr[ 4 ] = '2';     // ascii '2'
arr[ 5 ] = '3';     // ascii '3'
arr[ 6 ] = '4';     // ascii '4'
arr[ 7 ] = 0;       // integer 0

The loop prints for each successive value of i:

i == 0: "This "
i == 1: "is no"
i == 2: "t s"
i == 3: "o s"
i == 4: "imp"
i == 5: "le as"
i == 6: "it seems"

literally:

This is not so simple as it seems

The trailing \0 is to cause the while( arr[ i ] ) to fail and the loop to stop when i == 7.

Although this will probably cause the compiler to complain on non-microsoft compilers.