C isalpha() function
C isalpha(int ch)
The isalpha() function is used to check whether a character is an alphabet or not. The function is defined in the ctype.h header file.
Note: Letter (alphabet) - A letter is a segmental symbol of a phonemic writing system. The inventory of all letters forms an alphabet. Letters broadly correspond to phonemes in the spoken form of the language, although there is rarely a consistent and exact correspondence between letters and phonemes.
Syntax:
int isalpha(int argument);
isupper() Parameters:
Name | Description | Required /Optional |
---|---|---|
ch | ch is a character of class alpha in the current locale. | Required |
Return value from isalnum()
- The isalpha() function returns non-zero if ch is an alphabetic character; otherwise returns 0.
Example: C isalpha() function
#include <stdio.h>
#include <ctype.h>
int main()
{
char ch;
ch = 'A';
printf("\nIf %c is alphabet or not? %d",ch, isalpha(ch));
ch = '5';
printf("\nIf %c is alphabet or not? %d",ch, isalpha(ch));
ch = '+';
printf("\nIf %c is alphabet or not? %d",ch, isalpha(ch));
return 0;
}
Output:
If A is alphabet or not? 1 If 5 is alphabet or not? 0 If + is alphabet or not? 0
Example: C Program to test whether a character input from user is Alphabet or not
#include <stdio.h>
#include <ctype.h>
int main()
{
char ch;
printf("Input a character: ");
scanf("%c", &ch);
if (isalpha(ch) == 0)
printf("%c is not an alphabet.", ch);
else
printf("%c is an alphabet.", ch);
return 0;
}
Output:
Input a character: S S is an alphabet.
------------------------ Input a character: 9 9 is not an alphabet.
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C Programming: Tips of the Day
Reading a string with scanf :
An array "decays" into a pointer to its first element, so scanf("%s", string) is equivalent to scanf("%s", &string[0]). On the other hand, scanf("%s", &string) passes a pointer-to-char[256], but it points to the same place.
Then scanf, when processing the tail of its argument list, will try to pull out a char *. That's the Right Thing when you've passed in string or &string[0], but when you've passed in &string you're depending on something that the language standard doesn't guarantee, namely that the pointers &string and &string[0] -- pointers to objects of different types and sizes that start at the same place -- are represented the same way.
Ref : https://bit.ly/3pdEk6f
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