Condition to limit between 2 characters - c

Question

I'm writing code that need to limit the use to enter characters that be only from A to H. Greater then H should not be accepted.
I saw that with numbers I can use that like:
if (input == 0 - 9) return 1;
But, how I do that A to H (char)?

Answer 1

The C Standard does not specify that character encoding should be ASCII, though it is likely. Nonetheless, it is possible for the encoding to be other (EBCDIC, for example), and the characters of the Latin alphabet may not be encoded in a contiguous sequence. This would cause problems for solutions that compare char values directly.
One solution is to create a string that holds valid input characters, and to use strchr() to search for the input in this string in order to validate:
#include <stdio.h>
#include <string.h>
int main(void)
{
char *valid_input = "ABCDEFGH";
char input;
printf("Enter a letter from 'A' - 'H': ");
if (scanf("%c", &input) == 1) {
if (input == '\0' || strchr(valid_input, input) == NULL) {
printf("Input '%c' is invalid\n", input);
} else {
puts("Valid input");
}
}
return 0;
}
This approach is portable, though solutions which compare ASCII values are likely to work in practice. Note that in the original code that I posted, an edge case was missed, as pointed out by #chux. It is possible to enter a '\0' character from the keyboard (or to obtain one by other methods), and since a string contains the '\0' character, this would be accepted as valid input. I have updated the validation code to check for this condition.
Yet there is another advantage to using the above solution. Consider the following comparison-style code:
if (input >= 'A' || input <= 'H') {
puts("Valid input");
} else {
puts("Invalid input");
}
Now, suppose that conditions for valid input change, and the program must be modified. It is simpler to modify a validation string, for example to change to:
char *valid_input = "ABCDEFGHIJ";
With the comparison code, which may occur in more than one location, each comparison must be found in the code. But with the validation string, only one line of code needs to be found and modified.
Further, the validation string is simpler for more complex requirements. For example, if valid input is a character in the range 'A' - 'I' or a character in the range '0' - '9', the validation string can simply be changed to:
char *valid_input = "ABCDEFGHI0123456789";
The comparison method begins to look unwieldy:
if ((input >= 'A' && input <= 'I') || (input >= '0' && input <= '9')) {
puts("Valid input");
} else {
puts("Invalid input");
}
Do note that one of the few requirements placed on character encoding by the C Standard is that the characters '0', ..., '9' be encoded in a contiguous sequence. This does allow for portable direct comparison of decimal digit characters, and also for reliably finding the integer value associated with a decimal digit character through subtraction:
char ch = '3';
int num;
if (ch >= '0' && ch <= '9') {
printf("'%c' is a decimal digit\n", ch);
num = ch - '0';
printf("'%c' represents integer value %d\n", ch, num);
}

Answer 2

The if statement you present here is equal to:
if (input == -9) return 1;
which will return 1 in the case of an input equal to -9, so there is no range checking at all.
To allow numbers from 0 to 9 you have to compare like:
if (input >= 0 && input <= 9) /* range valid */
or with the characters that you want (A to H)1:
if (input >= 'A' && input <= 'H') /* range valid */
If you want to return 1 if the input is not in a valid range just put the logical not operator (!) in front of the condition:
if (!(input >= 'A' && input <= 'H')) return 1; /* range invalid */
1 You should take care of the used range if working with conditions that uses character ranges because the range needs an encoding that specify the letters in an incrementing order without any gaps in between the range (ASCII code e.g.: A = 65, B = 66, C = 67, ..., Z = 90).
There are encoding where this rule breaks. As the other answer of #DavidBowling stated there is for example EBCDIC (e.g.: A = 193, B = 194, ..., I = 200, J = 209, ..., Z = 233) which has some gaps in between the range from A to Z. Nevertheless the condition: (input >= 'A' && input <= 'H') will work with both encodings.
I never fall about such an implementation yet and it is very unlikely. Most implementations uses the ASCII code for which the condition works.
Nevertheless his answer provides a solution that is working in every case.

Answer 3

It's as simple as:
if(input >='A' && input<='H') return 1;
C doesn't let you specify ranges like 0 - 9.
In fact that's an arithmetic expression "zero minus nine" and evaluates to minus nine (of course).
Nerd Corner:
As others point out this is not guaranteed by the C standard because it doesn't specify a character encoding though in practice all modern platforms encode these characters the same as ASCII. So it's very unlikely you will come unstuck and if you're working in an environment where it won't work you'd have been told!
A truly portable implementation could be:
#include <string.h>//contains strchr()
const char* alpha="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const char* pos=strchr(alpha,input);
if(pos!=NULL&&(pos-alpha)<8) return 1;
This tries to find the character in an alphabet string then determines if the character (if any) pointed to is before 'I'.
This is total over engineering and not the answer you're looking for.