printing define value

I have written three programs first two program are working as it supposed to do but i don't understand what's happening in last program

Program 1

Code:

---

#include<stdio.h>


#define value  'A'


int main()
{
    printf(" %c \n", value );       
       
        return 0;
}

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A

Program 2

Code:

---

#include<stdio.h>


#define value  1


int main()
{
    printf(" %d \n", value );       
       
        return 0;
}

---

1


Last program

Code:

---

#include<stdio.h>

#define value  "A"


int main()
{
    printf(" %c \n", value );       
       
        return 0;
}

---

d



Why program print d instead of A? What does value store in a program A or d?

Maybe you can use %s instead of %c in the last case.

Or learn to use your compiler.

Code:

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$ gcc -Wall foo.c
foo.c: In function ‘main’:
foo.c:8:12: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Wformat=]
    printf(" %c \n", value );
            ^

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Quote:

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Originally Posted by Salem

Maybe you can use %s instead of %c in the last case.

Or learn to use your compiler.

Code:

---

$ gcc -Wall foo.c
foo.c: In function ‘main’:
foo.c:8:12: warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Wformat=]
    printf(" %c \n", value );
            ^

---

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Really I made silly mistake I didn't notice this is happening because of the format specifier