Need some assistance with understanding char->ASCII and some basics - c

Question

I had a project to write this program that reads a character, integer and a floating number. It also had to convert the character in to its ASCII integer. I vaguely understand most of it except the code regarding the ASCII number. I created most of it on my own but when it came to converting the character I had a buddy help me, but he is not the best at explaining this.
Any explanation would be very helpful, thank you. I did this all by trial and error for... probably far too long :P
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char c;
int i;
float f;
printf ("Please enter a character: ");
scanf ("%c", &c);
printf ("Please enter an integer: ");
scanf ("%d", &i);
printf ("Please enter a floating point number: ");
scanf ("%f", &f);
printf ("\n");
printf ("The character you entered is: %c which is %d in integer-specification\n", c,c);
printf ("The integer you entered is: %d.\n", i);
printf ("The floating-point number you entered is: %f.\n", f);
return 0;
}

Answer 1

It will be helpful to have a reference documentation of printf handy.
When the format specifier is %c, printf expects an int. A char works since it is promoted to an int.
When the format specifier is %d, printf expects an int. A char works since it is promoted to an int.
For the first case, printf prints the character corresponding to the value of the int.
For the second case, printf just prints the number.

Answer 2

Essentially a char is just an integer with a range from 0 to 255, the only thing that makes it a 'character' is that we have assigned characters to be represented by certain numbers.
What's going on in the printf() is that first you tell it that you're going to pass a char and would it please just output that byte to be displayed, then you say that you want it to convert an integer to a string and output that.
What it boils down to is that the ASCII code is the character.
Note: for reasons that are not important at the moment when being passed into a vararg function like printf() chars and shorts are promoted to ints.

Answer 3

Here is my attempt at explaining this.
Everything in a program memory is stored as sequences of 0s and 1s (binary numbers). You are probably familiar with bits and bytes, but just in case: a binary digit (0 or 1) is a bit. A sequence of 8 bits is a byte. Characters, such as 'A', 'b', '1', etc. have to be stored as numbers in memory. There are different ways to map a character to a number: EBCDIC (very rare these days, was used on large mainframes in the old days), UTF-8, UTF-16, UTF-32 (types of Unicode), ASCII. These mappings are called character encodings, and ASCII is the simplest of those commonly used.
So, when you enter a character, it gets stored in the variable c as an integer number encoded using ASCII. The content of memory location corresponding to the variable can be thus viewed both as a character and an integer.
When using the printf() function you can request the same variable to be printed using different possible representations. In this case, since c is a character variable and can be represented as a character or a number, you can use the different format specifiers, %c and %d, to have it printed as character and integer, respectively.
The 1st of the printf() calls above takes 3 arguments. The first argument is a string that is printed out, with things that start with % being used to interpret the remaining arguments. %c corresponds to the 2nd argument, %d to the 3rd. %c says: "treat the 2nd argument as character and print it out." %s says: "treat the 3rd argument as integer and print it out."
The other 2 printf() calls in your code take only 2 arguments. The 1st argument, the string, contains only one thing starting with % (format specifier), so we only need 1 additional argument to be interpreted using the format specifier.
Hope this is helpful.

Answer 4

the Man ascii will make you understand