Isalpha Function Issue

[mcertini]

In the below code, I am attempting to count the number of letters and number of words in a phrase. In researching the return values of isalpha(), I noted that when a letter is present the first character isalpha() output is 1. But the second character isalpha() output is 2? The phrase that was input into this example was the following: How do you organize? Can someone help me with why I am not getting uniform output for the "H" character and the "o" character?

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

int main(void)
	{
	int ch;
	float avg = 0;
	int spaces = 0;
	int letters = 0;
	int test = 0;
	int ltrcnt = 0;
	int spccnt = 0;		
	while((ch = getchar()) != '\n')
		{
		test = isalpha(ch);
		if(isalpha(ch) == 2)
			{
			ltrcnt++;
			}
		if(isspace(ch) == 8)
			{
			spccnt++;
			}
		}
	printf("The Number Of Words Are %d , The Number Of Letters Are %d ",(spccnt+1),ltrcnt);
	
	}

[laserlight]

In researching the return values of isalpha(), I noted that when a letter is present the first character isalpha() output is 1. But the second character isalpha() output is 2? The phrase that was input into this example was the following: How do you organize? Can someone help me with why I am not getting uniform output for the "H" character and the "o" character?

The return value of isalpha is either zero (false) or non-zero (true).

If you want to differentiate between uppercase and lowercase letters, use isupper and islower respectively.

[mcertini]

laserlight,

Thank you for your reply. As you state and what I have learned from another poster:

isalpha returns zero if the given character is not an alphabet.
Returns non-zero if it is alphabet.
And may return '1' if the alphabet is of upper case, and '2' if it is lower case.

[grumpy]

"another poster" has misinformed you.

There is no guarantee about what non-zero value is returned if a character is alphabetic. There is certainly no requirement to return a different result based on upper or lower case.

Some libraries might do that, but others will not.

[Elysia]

I should inquire as to whether you intend to use C or C++? What you have shown so far is C code.