Printing chars and their ASCII-code in C

Question

How do I print a char and its equivalent ASCII value in C?

Answer 1

This prints out all acsii values

void main()
{
int i;
i=0;
do
{
printf("%d %c \n",i,i);
i++;
}
while(i<=255);
}

and this prints out the acsii value for a given character:

void main()
{
int e;
char ch;
clrscr();
printf("\n Enter a character : ");
scanf("%c",ch);
e=ch;
printf("\n The ASCII value of the character is : %d",e);
getch();
}

Answer 2

Try this:

char c = 'a'; // or whatever your character is
printf("%c %d", c, c);

The %c is the format string for a single character, and %d for a digit/integer. By casting the char to an integer, you'll get the ascii value.

Answer 3

This reads a line of text from standard input and prints out the characters in the line and their ASCII codes:

#include <stdio.h>

void printChars(void)
{
    char    line[80+1];
    int     i;

    // Read a text line
    if (fgets(line, 80, stdin) == NULL)
        return;

    // Print the line chars
    for (i = 0;  line[i] != '\n';  i++)
    {
        int     ch;

        ch = line[i];
        printf("'%c' %3d 0x%02X\n", ch, ch, ch);
    }
}

Answer 4

This should work in any C system, not only those which are ASCII or UTF-8 based:

printf ( " Q is decimal 81 in ASCII\n" );

You did ask for a char; the other 96 of them are left as an exercise for the reader.