How to get the value associated to a char in c?-CodePudding

I would like to know how could I use the value of an input char in C.

Ex:

The user types a, how could I use the value of a (97) in my code? Is that possible?

MORE DETAILS: I made a while loop to get inputs from the user. If the user inputs a number 1-9, then I increment a variable called numbers, but the problem comes now: If the user types a for example, I made a condition to check wether the input is greater than 95 or not, and it does not work. What I want is to get this input (scanf("%d")) and if it is a letter (ascii >= 97), then increment another variable called letters.

Part of the code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    long long int teclas=0, letras=0, numeros=0, i=0, aux=0, k=0, j=0, flag=0,
    numprint=0, seq=0;
    scanf("%lld", &teclas);

    while(i<=teclas){
        scanf("%lld", &aux);
        if(aux>0 && aux<10){
            numeros  ;
            aux = 0;
        }else if(aux>96 && aux<123){
            letras  ;
            aux = 0;
        }
        i  ;
    }

CodePudding user response：

Yes, you can use the ASCII value of characters. Refer to the below example.

char c;
scanf("%c", &c);

int asciiValue;
asciiValue = c; // Here we are typecasting char to int implicitly. 
printf("ASCII val of %c is %d", c, asciiValue);

Here the variable asciiValue will have the int value of the character, and you can use it for any operations that you may want to perform in your code.

Note: char is 1 byte signed integer whereas int is a 4-byte signed integer.

CodePudding user response：

What I want is to get this input (scanf("%d")) and if it is a letter (ascii >= 97), then increment another variable called letters.

The short answer is: You can't read a letter using the format specifier %d

But you can do it in another way by checking the return value of scanf.

Something like:

int input_number;
char input_char;
if (scanf("%d", &input_number) == 1)  // Try to pass input as an int
{
    // The user typed a valid integer number which is now stored in input_number
    if(input_number>0 && input_number<10)
    {
        numeros  ;
    }
}
else if (scanf("%c", &input_char) == 1)  // Didn't get an int so read a single char
{
    // The user typed a non-integer char which is now stored in input_char
    //
    // Now you can do:
    if (input_char >= 'a' && input_char <= 'z')
    {
        letras  ;
    }
}
else
{
    // Input error
    exit(1);
}

CodePudding user response：

char c='a'
int value=a;
printf("%d",value);

This is called as implicit casting.output of this program will be 97 which is ascii code for a;

OR You can directly print it using

printf("%d",'a');

Output

CodePudding user response：

If you want the ASCII code then you can just do this:

#include <stdio.h>

int main()
{
    char a = 'a';
    int x = (int)a;
    printf("%d", x);
    return 0;
}

CodePudding user response：

You already get what you need: the value in the variable. Wheter it is a character('A','a') or an integrate number(65, 97) only depends on how you interpret it. See the example below:

#include <stdio.h>
#include <ctype.h>

int main(int argc, char **argv) {
  char input_buf = 0;

  printf("input any character:\n");
  while (1) {
    scanf("%c", &input_buf);
    if (isprint(input_buf)) {
      printf("\nthe value you just input: %c, the value: %d "
             ",\n Ctrl C or Ctrl Z to exit\n",
             input_buf, (int)input_buf);
      printf("\ninput any character:\n");
    }
  }

  return 0;
}