Thread: is there a weird rule about symbolic constants?

ok im new so if this question is wayy to easy to answer, forgive me but i have no idea....

the book im learning from says that symbolic constants are exact same values the thing they representing except you can use a different name to represent that value.. but how come this:

Code:

#include <stdio.h>

#define RESPONSE1 1
#define RESPONSE2 2


int main(void)
{
    int c;
    
    printf("press 1 for a response, 2 for a different one\n");

     while ((c = getchar()) != EOF){
        if(c == '1')
           printf("hehehhehe.....\n");
        if(c == '2')
           printf("fool...\n");
        }
}

i wrote this while i was bored but found out that "c == '1'" outputs printf but not "c == RESPONSE1"??? shouldnt it be the same thing??

c is an integer, and the vale of '1' and '2' is 31 and 32 in hex respectively. So, what you're wanting is:

Code:

if ( c == 1 ) // do stuff
if ( c == 2 ) // do stuff

Code:

and (c = getchar() - '0' )

to be coherent with logic.

no i meant that only c == '1' works, not c == 1 or c == RESPONSE1.

i even tried it out with a new program:

Code:

#include <stdio.h>

int main(void)
{
    int c;
    while((c = getchar()) != EOF)
    if(c == 2)
       printf("hehe\n");
}

even with this, i tried if(c == 2) and no printf shows, only c == '2'.... y is it like this?

2 is an integer value, unlike '2', which holds the integer value for the character '2'
so what you really want is to

#define RESPONSE1 '1'

you need to know that '1' is not equivalent to 1.

see this link: http://www.lookuptables.com/

'1' is a character represented in ansii and has a corresponding int value - 49. so c == '1' is same as c == 49. (referrring to link)

the reverse is true. so the int value 1 has a corresponding character - SOH

c changes all character constants to their int values.

therefore c= 'a' *2 is same as c=49*2.

this also answers your question why c == RESPONSE1 doesn't work.