how to check if string contains only alphabet

If we have a string like "CPROGRAMMING"
How can we check if this string contains only
1. Capital letters
2. No digits
Thanks

For starters:

isalpha

Are you familiar with looping through elements of an array?

Seeing what you have done so far would be nice.

isupper() would be handy too.

That is what i have just coded

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main ( int argc , char* argv[])
{
char* s = "CPROGRAMMING";
int i;
boolean flag = true;
for(i=0; , i< strlen(s) && flag == true; i++ )
{
   if( isalpha(s[i]) && isupper(s[i]) )
    flag = true;
  else
    flag = false;
}
if (flag)
    printf("A string contains only uppercase and alphabet letters\n" );
return 0;
}

---

The problem is at read line.isupper() and isalpha() take arguments as integer type however like you said we are passing arg like char type. That is why i am confused

Just pass the char as an integer. I'm not sure why they use integers for those functions. It's not like they'll except -1 as an argument. But they expect a regular character in the range 0 to 127.

Not giving away too much you could then also do:

Code:

---


...
while (*str++)
    {
      if ((*str >= 'a') && (*str <= 'z'))
...

---

But yes, those functions will take your passed in char data type as well.

>I'm not sure why they use integers for those functions.
>It's not like they'll except -1 as an argument.
Well, technically... Those functions accept int to account for EOF, the idea being that any return value from getchar is an acceptable input to any of the character classification functions (ie. the range of unsigned char, plus EOF). And EOF is often implemented as -1. :D

>But they expect a regular character in the range 0 to 127.
Beware giving out specific ranges, especially when the range of a type is implementation-defined.

@slingerland3g: >='a' and <='z' can help me.
@King Mir : what do you mean pass char as an integer. Could you explain pls
Thanks all

Thungmail's code is an example. s is a char * and s[i] is therefor a char.

Only s should really be declared as a const char pointer, as it points to a literal string which cannot be changed.

Quote:

---

Originally Posted by King Mir

Just pass the char as an integer. I'm not sure why they use integers for those functions. It's not like they'll except -1 as an argument. But they expect a regular character in the range 0 to 127.

---

The isXXX() functions are required to accept EOF. EOF, by definition, is outside of the range 0..255. Therefore, a char is not big enough.

(EOF may or may not be equal to -1 -- the point is, it is not representable in a char)

Quote:

---

Originally Posted by thungmail

That is what i have just coded

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main ( int argc , char* argv[])
{
char* s = "CPROGRAMMING";
int i;
boolean flag = true;
for(i=0; , i< strlen(s) && flag == true; i++ )
{
   if( isalpha(s[i]) && isupper(s[i]) )
    flag = true;
  else
    flag = false;
}
if (flag)
    printf("A string contains only uppercase and alphabet letters\n" );
return 0;
}

---

The problem is at read line.isupper() and isalpha() take arguments as integer type however like you said we are passing arg like char type. That is why i am confused

---

Among the other reasons, try this code to get an hint:

Code:

---

#include<stdio.h>

int main(void)
{
        printf("sizeof('a') = %d\n", (int)sizeof('a'));
        return 0;
}

---

P.S.
Notice instead that if you try to compile it as c++ code, you'll get that sizeof('a') = 1.