Storing ASCII code point value in a float prevents its interpretation as a character (in C)

Question

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.

Answer 1

You asked to write the first byte of the float. That does not produce anything useful.

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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.

• The char ninety-seven is represented by byte 61 (hex).^[1]
• The int ninety-seven is represented by bytes 61 00 00 00 or 61 00 00 00 00 00 00 00.^[1]
• The 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.

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1. I'm assuming a modern desktop system.