Nibble half digit converter

Question

Hello fellow Programmers,

I might want to ask you ,

I`ve got a task from my University to convert a decimal digit (0-9) to a half bit (nibble = 0101 ie.) the restriction is that we cannot use if statements

so i tried a little bit and came so far

int main()
{
    int n;

    printf("type in your digit");
    scanf("%d", &n );
    printf("%n%n%n%n", (n/8)%2, (n/4)%2, (n/2)%2, n%2);

    return 0;
}

My problem here is that i cant put in a digit if i do so the program crashes

im really struggling and dont know how to proceed or make it work

can you give me some examples of how it could work or some good advice how to make it work.

i would really appreciate it

Thank you really for any help and answer!

`%n` doesn't do what you think it does. Read the documentation for `printf`. — user3386109, Apr 26 '16 at 21:00
to convert a ascii digit to a nibble use: `char nibble = digit - '0';` the resulting value in 'nibble' will be in the range 0x00 through 0x09. To print individual bits: `printf( "%i%i%i%i", ((nibble>>3)&1)+'0', ((nibble>>2)&1)+'0', ((nibble>>1)&1)+'0', (nibble&1)+'0' );` — user3629249, Apr 28 '16 at 14:03

score 0 · Accepted Answer · answered Apr 26 '16 at 21:01

You're using the wrong format specifier to printf.

The %n format specifier expects an int * and writes the number of characters written so far to the given memory address. Because you pass in a int to an argument expecting an int *, printf interprets that number as a memory address and tries to write to it. This is undefined behavior, and will most likely result in a core dump.

You want to use %d instead, is used to output an int:

printf("%d%d%d%d", (n/8)%2, (n/4)%2, (n/2)%2, n%2);

Nibble half digit converter

1 Answers1