Question

unsigned char ascii;
int a = 0;
char string[4] = "foo"; "A1"
ascii = (string[a] - 'A' + 10) * 16;

warning: conversion to ‘unsigned char’
from ‘int’ may alter its value

It seems that gcc considers chars and number literals as int by default. I know I could just cast the expression to (unsigned char) but how can I specify char literals and number literals as 8 bit without casts ?

A similar issue:

Literal fractions are considered double by default but they can be specified to float by:

3.1f

Therefore, 3.1 would be considered a float rather than a double.

Answer 1

In C, you cannot do calculations in anything shorter than int

char a = '8' - '0'; /* '8' is int */
char a = (char)'8' - '0'; /* (char)'8' is converted to `int` before the subtraction */
char a = (char)'8' - (char)'0'; /* both (char)'8' and (char)'0' are converted */