How to print ASCII value of 2 characters? - c

Question

I'm trying to print ASCII value of "\t" i.e escape sequence.
But my program only prints the ASCII value of "\" i.e ASCII value will be 92.
Is there any way to print ASCII value of 2 characters ?
Help would be really appreciated.I have included my code below.
#include<stdio.h>
main()
{
char b=0;
printf("Enter any character to print it's ASCII value : ");
scanf("%c",&b);
printf("The ASCII value of '%c' is %d",b,b);
return 0;
}

Answer 1

Capture input of backslash and handle it as a separate input. A switch can be used to print the result for the escaped characters.
#include <stdio.h>
int main( void) {
char b=0;
printf("Enter any character to print it's ASCII value : ");
if ( 1 == scanf(" %c",&b)) {
if ( b == '\\') {//read a backslash
if ( 1 == scanf(" %c",&b)) {
switch ( b) {
case 't' :
printf("The ASCII value of '\\t' is %d\n", '\t');
break;
case '\\' :
printf("The ASCII value of '\\' is %d\n", '\\');
break;
default :
printf ( "not handled yet\n");
}
}
}
else {
printf("The ASCII value of '%c' is %d\n",b,b);
}
}
return 0;
}
Output
Enter any character to print it's ASCII value : \t
The ASCII value of '\t' is 9
Enter any character to print it's ASCII value : \\
The ASCII value of '\' is 92
To get the code for a backslash, two backslashes must be entered

Answer 2

This code might help you understand what is going on here:
#include <stdio.h>
const char* escaped(int ch) {
switch (ch) {
case '\t': return "tab";
case '\n': return "newline";
case '\r': return "carriage return";
// continue with other escaped characters
default: return "not escaped";
}
}
int main()
{
char b = 0;
printf("Enter any character to print it's ASCII value : ");
scanf("%c", &b);
printf("The ASCII value of '%c' is %d and is also known as: %s\n", b, b, escaped(b));
}
Just to be really clear, for a tab, at the keyboard you just press the tab key. You do not enter the string "\t". The string "\t" will be interpreted as 2 characters: '\' and 't'.
The \ escape code is something you would use in writing strings in your C code.
For example, if you type the string "trying" into some C source code, then you are entering the stream of characters: t r y i n g but if you type the string: "\trying" then the first character is indicating that this is an escaped character and is a convenient way to indicate that you really wanted a tab followed by the characters: r y i n g.
With the above code, if you enter "\t" scanf is only getting from stdin, one character at a time, so it just takes the first character, the backslash (start of an escape sequence), and will print out:
Enter any character to print it's ASCII value : \t
The ASCII value of '\' is 92 and is also known as: not escaped
the 't' character is still in the input stream and the program has not handled this character.
scanf does not interpret "\t" as a tab but as the character stream \ followed by t. That is causing your confusion.
You will experience some problems in running this code. For example, you want to try backspace, which is \b or decimal 8 value. scanf will probably ignore it and assume you are attempting to backspace a previous character. But for other escaped characters such as tab character and newline it will work.
Have a read of this: https://en.wikipedia.org/wiki/Escape_character

Answer 3

#define INV (EOF -1)
int z = 0, tmp = INV;
char buff[6] = "\\0";
while (1)
{
if (tmp == INV)
z = getchar();
else
{
z = tmp;
tmp = INV;
}
if (z == EOF || tmp == EOF) break;
if (z == '\\')
{
tmp = getchar();
switch (tmp)
{
case 't':
z = '\t';
tmp = INV;
break;
case 'b':
z = '\b';
tmp = INV;
break;
case 'r':
z = '\r';
tmp = INV;
break;
case 'n':
z = '\n';
tmp = INV;
break;
// add here anothere ones
default:
break;
}
}
printf("Char \\0%03o - '%c'\t\t has the ascii code of %03d decimal 0x%02X hexadecimal\n", z, z < 32 ? ' ' : z, z, z);
}