In C is there a way to convert an ASCII value typed as an int into the the corresponding ASCII character as a char?
You can assign int to char directly.
int a = 65;
char c = a;
printf("%c", c);
In fact this will also work.
printf("%c", a); // assuming a is in valid range
If i is the int, then
char c = i;
makes it a char. You might want to add a check that the value is <128 if it comes from an untrusted source. This is best done with isascii from <ctype.h>, if available on your system (see #Steve Jessop's comment to this answer).
If the number is stored in a string (which it would be if typed by a user), you can use atoi() to convert it to an integer.
An integer can be assigned directly to a character. A character is different mostly just because how it is interpreted and used.
char c = atoi("61");
char A;
printf("ASCII value of %c = %d", c, c);
In this program, the user is asked to enter a character. The character is stored in variable c.
When %d format string is used, 65 (the ASCII value of A) is displayed.
When %c format string is used, A itself is displayed.
Output:
ASCII value of A = 65
#include <stdio.h>
#include <cs50.h>
int d;
int main(void){
string a = get_string("enter your word: ");
int i = 0;
while(a[i] != '\0'){
printf("%i ", a[i]);
i++;
}
printf("\n");
}
Related
If you run the following code you will get the ASCII characters, even though we use int i within %c inside the for loop.
#include <stdio.h>
int main()
{
int i;
for (i = 0; i <= 255; i++) /*ASCII values ranges from 0-255*/
{
printf("ASCII value of character %c = %d\n", i, i);
}
return 0;
}
Could you please advise how is this possible since there are characters inside ASCII?
Every variable in the computer is stored as binary and its value will depend on the type you define. For example, in memory store the value 1010. If you cast it as int, it will print -6, if you cast it as unsigned int, it will print 10. Same for int and char. It depends on you
printf() - Print formatted
we use format specifiers to specify the format of given variable,
%d format specifier to display the value of an integer variable.
%c is used to display character
as ASCII values ranges from 0-256 , %c will print the respective character symbol of number in Integer i
I'm trying to print out the digits of an integer using the char format specifier.
This is what I currently have
//counter = arbitrary int
while (counter > 0) {
char c = (char)(counter % 10);
printf("%c", c);
counter/=10;
}
This just prints out blank but if I change the format specifier to int but leave the type cast, it'll print the correct value. Something like this
char c = (char)(counter % 10);
printf("%d", c);
Any ideas why this isn't printing when I use the char format specifier?
%c is for printing a character, not a number. If you want to print a number, you should use %d. The reason it works for char is because char is implicitly promoted to int when you pass it as an argument to a variadic function.
If c contained 'A', and you used %c to format it, you would get the letter ‘A’, but if you used %d, you would get its integer value, e.g. 65.
If you absolutely must use %c, you can add the value of '0' to the character before printing, e.g.L
printf("%c", c + '0');
%c can be used in this context as printf("%c",c+'0') where +'0' will turn it into the integer equivalent of c.
what happens if I read an integer like 20,30,10000...9999 into variable a ? it only prints the first digit in the number that I've read...why is that ?
for example if I read 123, on the screen it prints 1. Isn't it supposed to convert the integer 123 into it's equivalent ASCII character representation ?
#include <stdio.h>
int main() {
char a;
scanf("%c", &a);
printf("%c", a);
return 0;
}
This is an exam question from C language.
No, it reads the character, which is represented by the machine as a small integer, into the variable.
If you enter 100 (the number 100, three keypresses and thus three characters), it will only store the first character of that, i.e. the leading 1.
If you wanted to convert a number to an actual integer, you should use %d and an int variable of course.
Printing with %c will print back a single character, by interpreting the small integer value as a character (rather than as an integer). So for an input of 100 you will see 1 printed back out, i.e. the character that represents the decimal digit one.
If you want to print out the numeric representation of the character you read in, scan with %c but print with %d, and cast the char to (int) in the printf() call.
The problem is that %c parse a char for console input. From a number like 123 it take only the first letter and dispose the rest. The way to parse a int value is using %d on the scanf function.
No, it will read only the first character into the char variable. How can a char variable store more than one character at an instant? It can't.
So if you want the ASCII value, input as an integer instead.
int a;
scanf("%d", &a); // suppose input is 65
printf("%c", a); // prints 'A'
printf("%d", a); // prints 65
Whereas
char a;
scanf("%c", &a); // suppose input is 65
printf("%c", a); // prints '6'
printf("%d", a); // prints 54 which is the ASCII value of '6'
I need to take a keyboard input letter, and save that letter's decimal value as an integer.
How can I do that with scanf ?
I just figured out myself.
if I want to store the decimal value of 'z'
I can just do int value='z'-'a'+ 97
This is what I wanted to know.
This code reads a char from keyboard (stdin) using scanf(), stores it in byte-sized variable c of type char, and then prints its ASCII decimal value as an int to stdout :
char c;
scanf("%c", &c);
printf("%d", c);
If you want to output the ASCII value, simply printf() with the %d format string.
char ch = 'a';
printf("%d", a); // should be 97
This question already has answers here:
Printing chars and their ASCII-code in C
(8 answers)
Closed 5 years ago.
I have accepted a character as an input from the user. I want to print the ASCII value of that character as an output. How can I do that without using any pre-defined function (if it exists) for the same?
Instead of printf("%c", my_char), use %d to print the numeric (ASCII) value.
Also consider printf("%hhu", c); to precisely specify conversion to unsigned char and printing of its decimal value.
Update0
So I've actually tested this on my C compiler to see what's going on, the results are interesting:
char c = '\xff';
printf("%c\n", c);
printf("%u\n", c);
printf("%d\n", c);
printf("%hhu\n", c);
This is what is printed:
� (printed as ASCII)
4294967295 (sign extended to unsigned int)
-1 (sign extended to int)
255 (handled correctly)
Thanks caf for pointing out that the types may be promoted in unexpected ways (which they evidently are for the %d and %u cases). Furthermore it appears the %hhu case is casting back to a char unsigned, probably trimming the sign extensions off.
This demo shows the basic idea:
#include <stdio.h>
int main()
{
char a = 0;
scanf("%c",&a);
printf("\nASCII of %c is %i\n", a, a);
return 0;
}
The code printf("%c = %d\n", n, n); displays the character and its ASCII.
This will generate a list of all ASCII characters and print it's numerical value.
#include <stdio.h>
#define N 127
int main()
{
int n;
int c;
for (n=32; n<=N; n++) {
printf("%c = %d\n", n, n);
}
return 0;
}
#include "stdio.h"
#include "conio.h"
//this R.M.VIVEK coding for no.of ascii values display and particular are print
void main()
{
int rmv,vivek;
clrscr();
for(rmv=0;rmv<=256;rmv++)
{
if(printf("%d = %c",rmv,rmv))
}
printf("Do you like particular ascii value\n enter the 0 to 256 number");
scanf("%d",&vivek);
printf("\nthe rm vivek ascii value is=%d",vivek);
getch();
}