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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();
}
Related
When you declare two variables char a,b; and then you use first 'a' and then 'b',it prints only b, but if you declare it 'b' then 'a', it has no problem printing both in ASCII,the point of the program is to read 121 and 120 and to print yx. the problem - https://prnt.sc/pr5nww
and if you swap them -https://prnt.sc/pr5mt5
#include <stdio.h>
#include <stdlib.h>
int main(){
char a,b;
scanf("%d",&a);
scanf("%d",&b);
printf("%c",a);
printf("%c",b);
}
This is kind of a confusing situation. When it comes to mixing char and int values (as you might do when investigating the numeric values of characters in a character set), it turns out the rules for scanf and printf are almost completely different.
First let's look at the scanf lines:
char a,b;
scanf("%d",&a);
scanf("%d",&b);
This is, in a word, wrong. The %d format in scanf is for scanning int values only. You cannot use %d to input a value of type char. If you want to input a character, the format for that is %c (although it'll input it as a character, not a number).
So you'd need to change this to
char a,b;
scanf("%c",&a);
scanf("%c",&b);
Now you can type characters like A and $ and 3 and have them read into your char variables a and b. (Actually, you're going to have additional problems if you hit the Return key between typing the characters for a and b, but that's a different story.)
When it comes to printing the characters out, you have a little more freedom. Your lines
printf("%c",a);
printf("%c",b);
are fine. And if you wanted to see the integer character-set values associated with the characters, you could have typed
printf("%d",a);
printf("%d",b);
and that would have worked, too. This is because when you call printf (and other functions ike it), there are some automatic conversions that take place: types char and short int are automatically promoted to (passed as) int, and type float is promoted to double. But these automatic conversions happen only for values of those types (as when calling printf). There a=is no such conversion when you're passing pointers to these types, as when calling scanf.
What if you wanted to read numbers, not characters? That is, what if you wanted to input the number 65 and see it get printed as capital A? There are several possible ways to do that.
The first way would be to continue to use %d in your scanf call, but change the type of your variables to int:
int a,b;
scanf("%d",&a);
scanf("%d",&b);
Now you can print a and b out using either %c or %d, and it'll work fine.
You could also use a temporary int variable, before reassigning to char, like this:
char a,b;
int tmp
scanf("%d",&tmp);
a = tmp;
scanf("%d",&tmp);
b = tmp;
The final, lesser-known and somewhat more obscure way, is to use the h modifier. If you say
char a,b;
scanf("%hhd",&a);
scanf("%hhd",&b);
now you're telling scanf, "I want to read decimal digits, but the target variable is a char, not an int."
And, again, you can print a and b out using either %c or %d, and it'll work fine.
the point of the program is to read 121 and 120 and to print yx
Do
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char a, b;
/* Scan into the half of the half of an int (the leading blank
makes scanf() eat whitespaces): */
scanf(" %hhd", &a);
scanf(" %hhd", &b);
/* Print the half of the half of an int: */
printf("%hhd", a);
printf("%hhd", b);
}
To print the characters literally do the printing part like this:
...
printf("%c", a);
printf("%c", b);
}
#include <stdio.h>
int main ()
{
double a=0;
char b=0;
scanf ("%d%c",&a,&b);
printf ("%d,%c", a, b);
return 0;
}
This is my code for a quick test program I wrote to play around with the scanf function in C. I am trying to have the user input something like 78X + 5 = 19 (then hit enter) and then parse that into variables a, b, and c where in this case a=78, b=5, c=19. In the sample code, when I type in 78X, c doesn't store a value to b and only prints "78, " and then terminates. Why won't it store a value to b?
If your input is 75x then below is the code which reads the value and stores it in a(75) and b(x) respectively
#include <stdio.h>
int main ()
{
int a=0;
char b=0;
scanf ("%d%c",&a,&b);
printf ("%d%c", a, b);
return 0;
}
The , in your format string is significant. The string %d,%c would match the input 78,x but it would not match 78x .
Also you need to use %f to scan and print a double. Using %d causes undefined behaviour (which may manifest itself as b seeming to not appear). Either change to %f, or change your double to an int.
Take for example
int i=10,j;
float b=3.14,c;
char str[30];
sprintf(str,"%d%f",i,b);
sscanf(str,"%d%f",&j,&c);
printf("%d ----- %f\n",j,c);
OUTPUT :- 103 ----- 0.1400000
As you can see, initially i=10 and b=3.14.
I want that j=10 and c=3.14 by using sprint() and sscanf().
The problem I am facing is that the compiler assigns j=103 and c=0.140000.
Is there any way to get rid of this problem in sscanf()?
Add one space to sprintf. Change:
sprintf(str,"%d%f",i,b)
to
sprintf(str,"%d %f",i,b)
Aside: It would be also safer to use snprintf here:
snprintf(str, sizeof str, "%d %f", i, b)
The best way will be to separate the numbers, using a different sign, but if you know that the first int is 2 chars long you can specify it:
sscanf(str,"%2d%f",&j,&c);
// ^^
You are missing a space
Change
sprintf(str,"%d%f",i,b);
to
sprintf(str,"%d %f",i,b);
#include <stdio.h>
int main()
{
int i=10,j;
float b=3.14,c;
char str[30];
sprintf(str,"%d %f",i,b);
sscanf(str,"%d%f",&j,&c);
printf("%d ----- %f\n",j,c);
return 0;
}
output
~ > ./a.out
10 ----- 3.140000
The %d conversion specifier in sscanf will match any number of contiguous numeric characters in the buffer pointed to by str till it encounters a non-numeric character which it can't match. This will cause the integral part of the float to be read as part of the int value. Therefore you must have a way to separate where you integer ends and float starts in the string str. You can put any non-numeric character as a sentinel to separate int value from the float value.
int i = 10, j;
float b = 3.14, c;
char str[30];
// a # to separate the two values, can be any non-numeric char so that it
// is not mistaken for a digit in the int or float value
sprintf(str,"%d#%f", i, b);
// match the separator character to read int and then float
sscanf(str, "%d#%f", &j, &c);
I'm trying to write a short program were:
#include <stdio.h>
void main()
{
char=a,b,c;
printf("please place 3 numbers:\n");
scanf("%c%c%c", &a,&b,&c);
}
The exercise I'm trying to solve is how to change the char to int so if I write in a the number 3, I will get the number 3 Printed.
at this point I'm only getting the value.
I would appreciate any help.
The answer depends somewhat on what you can assume about the character set. If it's something like ASCII (or really, any character set that includes the digits in sequential order), you just need to offset the character value by the value of the character 0:
int aValue = a - '0';
I'm sure that C# provides better ways to do what you're trying to do, though. For example, see this question for some examples of converting strings to integer values.
First of all your syntax need some checking
You should know that you declare a variable this way (a char in this example):
char a;
If you want to declare multiple variables of the same type in a row you do :
char a, b, c;
If you want to assign a value to a declared variable :
a = '3';
Now to print a char using printf (man printf is a must read, more infos are in coreutils) :
printf("%c", a);
If you want to get the char from the command line, I recommand you to use getchar() (man getchar) instead of scanf because if suits better what you are trying to achieve and doesn't require you to use a syntax in scanf that I am sure you don't fully understand yet.
Your question is incredibly light on details, so here are several options:
#include <stdio.h>
int main()
{
char a,b,c;
printf("please place 3 numbers:\n");
scanf("%c%c%c", &a,&b,&c);
printf("Printing ints (auto-promotion): %d %d %d\n", a, b, c);
printf("Printing ints (explicit-promotion): %d %d %d\n", (int)a, (int)b, (int)c);
printf("Printing digits: %d %d %d\n", a-0x30, b-0x30, c-0x30);
return 0;
}
If the input is 123,
I expect the output to be:
Printing ints (auto-promotion): 49 50 51
Printing ints (explicit-promotion): 49 50 51
Printing digits: 1 2 3
Some things I fixed along the way.
main should return an int, not be void.
char=a,b,c; is a syntax error. You meant char a,b,c;
added a return 0; at the end of main.
You question is not quite understandable. Still I'll try to help. I think that what you want is to store an integer value in the char variable. You can do so by using the following code:
#include<stdio.h>
void main()
{
char a,b,c;
printf("Enter three numbers:\n");
scanf(" %c %c %c",&a,&b,&c); //notice the spaces between %c
}
Or if you want to enter a character and print its ASCII value, you can use the following code:
#include<stdio.h>
#include<conio.h>
void main()
{
char a,b,c;
printf("Enter three characters:\n");
scanf(" %c %c %c",&a,&b,&c);
printf("Entered values: %d %d %d",a,b,c);
getch();
}
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");
}