float c;
c = 'a';
while (c <= 'z')
{
write(1, &c, 1);
c++;
}
write() is data type-agnostic. It should deliver the correct code point values to the terminal. The terminal should then interprete it according to its character encoding scheme (usually UTF-8).
Question Why does this code output a blank line rather than the alphabet? (the goal of this question is to understand the underlying mechanism at work)
What I have checked I printed out the float values of each incremented c. I seemingly always get the correct code point value ( 116.0000000 for 't', 117.0000000 for 'u' etc) I suspect the problem arises from storing the code point values as floats, potentially causing precision loss somehow.
You asked to write the first byte of the float. That does not produce anything useful.
Let's look at the following program instead:
char c = 'a';
write(1, &c, 1);
int i = 'a';
write(1, &i, 1);
float f = 'a';
write(1, &f, 1);
First of all, 'a' is an int, and it's equivalent to 97.[1]
c, i and f will be all be initialized to have the value ninety-seven, but that is represented differently by different types.
char ninety-seven is represented by byte 61 (hex).[1]int ninety-seven is represented by bytes 61 00 00 00 or 61 00 00 00 00 00 00 00.[1]float ninety-seven is represented by bytes 00 00 C2 42.[1]The subsequent calls to write output the first of these bytes.
61 for the char. Your terminal displays this as an a.[1]61 for the int. Your terminal displays this as an a.[1]00 for the float. Your terminal might display this as an empty space.