The code should accept a character and it should check whether its between 'j' and 'p'.

If it is between 'j' and 'p' it should print yes or else it should print no.

I have tried to do something about it but the only ideas I got is this:

if (a=='j' || a=='k' || a=='k' || a=='l' || a=='m' || a=='n' || a=='o' || a=='p')
{
   printf("YES");
} 
else
{
   printf("NO");
}

CodePudding user response：

You can avoid all the alternative tests by using a function like strchr():

if (strchr("jklmnop", a)) {
    puts("YES");
} else {
   puts("NO");
}

The obvious approach is to do something like

if (a >= 'j' && a <= 'p') {
    // ...
}

but that has a problem if you want to write portable code.

The C standard only requires that the characters '0' through '9' appear consecutively and in order. If you're following the standard to a t, you shouldn't assume that 'j' through 'p' appear together and can be used with a pair of >= and <= tests. If you add additional qualifications like requiring an ASCII compatible character set, it's a different story.

CodePudding user response：

It depends on what you mean by "between j and p".

If you mean "Only lowercase English letters between j and p", then one portable way of writing it down is

if (strchr("jklmnop", a)) ...

If you mean "Character codes between that of 'j' and that of 'p', in whatever encoding is used by the machine", then one portable way of writing it down is

if (a >= 'j' && a <= 'p') ...

If your encoding is ASCII, then the two notions above strictly coincide for any range of English letters.

If your encoding is EBCDIC, then they coincide for the range j..p, but not say for the range i..p.

It is guaranteed that all English letters between j and p are included in the range of the codes in any standards-compliant encoding, but there might be additional, non-English-letter characters in the same range.

Finally, for completeness, if by "between j and p" you mean "letters of the user's language, whatever it is, that are between j and p", then one correct way of writing it down is probably

setlocale (LC_ALL, ""); // first statement of the program

...
if (strcoll(a, "j") >= 0 && strcoll(a, "p") <= 0) ...

Note that here, a is not a character as above, but a string. It is up to you to ensure that it contains a single character of the user's language (which is not the same thing as a single char element). Ensuring this is very non-trivial.

TL;DR

if (a >= 'j' && a <= 'p') will probably work for whatever task you currently have, but don't assume it will always work.

CodePudding user response：

try this code

#include <stdio.h>

int main()
{
    char a;
    printf("enter the letter :");
    scanf("%c",&a);
    if(a>='j'  && a<='p')
    printf("yes the letter is between j and p");
    else
    printf("No the given letter is not between j and p");
}

CodePudding user response：

Each and every alphabet has an ASCII code which is an integer so you can perform it like this,

char a;
scanf("%c",&a);
if(a>='j'&&a<='p')
{
    printf("YES");
}
else
{
    printf("NO");
}

CodePudding user response：

Assuming lowercase letters form a contiguous block in the execution set, which is true for ASCII, you can write:

if (a >= 'j' && a <= 'p') ...

Assuming a is an int containing a char value, You can write this as a single test, but a good compiler should be able to generate the same code for the more readable test above:

if ((unsigned)(a - 'j') <= (unsigned)('p' - 'j')) ...

You could also test if a is in a set of characters, which will work regardless of the target encoding:

if (a != 0 && strchr("jklmnop", a)) ...

The test for a != 0 can be removed if you know a cannot be a null byte.

CodePudding user response：

Character literals (This is a character literal: 'a') are just numbers. Almost all computers use a ASCII-compatible encoding (there are very few exceptions). ASCII assumed, for example 'a' is a 97 for your computer, 'j' a 106. if you write a=='j' you basically write a==106. Using character literals is just syntax sugar, it makes it a lot easier for humans to read but the computer does not care.

This means, you have to check if a is between 106 and 112. You probably know a better way to do that than you current way. But instead of 106 and 112 write 'j' and 'p', because it easier to read.