Why Do We Add a char (A Byte) While We Need 3 Bytes?

Question

In copy.c, in case we're copying a 3x3 bmp image, this piece of code adds the calculated padding to the bmp

// then add it back (to demonstrate how)
for (int k = 0; k < padding; k++)
{
    fputc(0x00, outptr);
}

I believe that we're adding a hexadecimal value (3 Bytes), but per the fputc() man page

DESCRIPTION
fputc() writes the character c, cast to an unsigned char, to stream.

Now, the size of an unsigned char is a byte. So how does fputc() manage to put 3 Bytes?

Answer 1

What it does, Kareem, is it puts one byte with every iteration of the loop. padding is equal to 3 (i.e., 3 bytes) in this case! :)

Answer 2

fputc(0x00, output); really puts a single char.

0x00 in hexadecimal is exactly one byte. the 0x is not part of the number, it just indicate that the number is in fact hexadecimal.

Would be good clarify that, the padding in the example image is really == 1 (1 byte), at the end of each scanline. Since there are 3 vertical lines so there are a total of 3 bytes, but they are not written together in the same line.

If the image was 2x3, there would be padding == 2, per line. With a total of 6.