Check if User Inputs a Letter or Number in C - c

Question

Is there an easy way to call a C script to see if the user inputs a letter from the English alphabet? I'm thinking something like this:
if (variable == a - z) {printf("You entered a letter! You must enter a number!");} else (//do something}
I want to check to make sure the user does not enter a letter, but enters a number instead. Wondering if there is an easy way to pull every letter without manually typing in each letter of the alphabet :)

Answer 1

It's best to test for decimal numeric digits themselves instead of letters. isdigit.
#include <ctype.h>
if(isdigit(variable))
{
//valid input
}
else
{
//invalid input
}

Answer 2

#include <ctype.h>
if (isalpha(variable)) { ... }

Answer 3

isalpha() will test one character at a time. If the user input a number like 23A4, then you want to test every letter. You can use this:
bool isNumber(char *input) {
for (i = 0; input[i] != '\0'; i++)
if (isalpha(input[i]))
return false;
return true;
}
// accept and check
scanf("%s", input); // where input is a pointer to a char with memory allocated
if (isNumber(input)) {
number = atoi(input);
// rest of the code
}
I agree that atoi() is not thread safe and a deprecated function. You can write another simple function in place of that.

Answer 4

Aside from the isalpha function, you can do it like this:
char vrbl;
if ((vrbl >= 'a' && vrbl <= 'z') || (vrbl >= 'A' && vrbl <= 'Z'))
{
printf("You entered a letter! You must enter a number!");
}

Answer 5

The strto*() library functions come in handy here:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#define SIZE ...
int main(void)
{
char buffer[SIZE];
printf("Gimme an integer value: ");
fflush(stdout);
if (fgets(buffer, sizeof buffer, stdin))
{
long value;
char *check;
/**
* strtol() scans the string and converts it to the equivalent
* integer value. check will point to the first character
* in the buffer that isn't part of a valid integer constant;
* e.g., if you type in "12W", check will point to 'W'.
*
* If check points to something other than whitespace or a 0
* terminator, then the input string is not a valid integer.
*/
value = strtol(buffer, &check, 0);
if (!isspace(*check) && *check != 0)
{
printf("%s is not a valid integer\n", buffer);
}
}
return 0;
}

Answer 6

You can also do it with few simple conditions to check whether a character is alphabet or not
if((ch>='a' && ch<='z') || (ch>='A' && ch<='Z'))
{
printf("Alphabet");
}
Or you can also use ASCII values
if((ch>=97 && ch<=122) || (ch>=65 && ch<=90))
{
printf("Alphabet");
}

Answer 7

int strOnlyNumbers(char *str)
{
char current_character;
/* While current_character isn't null */
while(current_character = *str)
{
if(
(current_character < '0')
||
(current_character > '9')
)
{
return 0;
}
else
{
++str;
}
}
return 1;
}

Answer 8

You can implement the following function that returns a boolean, it checks whether the input is only composed by characters and not numbers, it also ignores spaces. Note that it supposes that the input in collected by fgets and not scanf. You should only change the while condition if you want to use another input method.
bool is_character(char text[])
{
bool just_letters;
just_letters = true;
while((text[i] != '\n') && (just_letters == true))
{
if ((text[i] >= 'A' && text[i] <= 'Z') || (text[i] >= 'a' && text[i] <= 'z') || text[i] == 32)
{
just_letters = true;
}
else
{
just_letters = false;
}
}
return just_letters;
}