I was reading some code from a book, when I decided to make a change to see what the uninitialized value of sec would be before the while statement:
#include<stdio.h>
#define S_TO_M 60
int main(void)
{
    int sec,min,left;
    printf("This program converts seconds to minutes and ");
    printf("seconds. \n");
    printf("Just enter the number of seconds. \n");
    printf("Enter 0 to end the program. \n");
    printf("sec = %d\n",sec);
    while(sec > 0)
    {
        scanf("%d",&sec);
        min = sec/S_TO_M;
        left = sec % S_TO_M;
        printf("%d sec is %d min, %d sec. \n",sec,min,left);
        printf("Next input?\n");
    }
    printf("Bye!\n");
    return 0;
}
This compiles under GCC with no warnings, even though sec is uninitialized at that point, and I get a value of 32767:
$ gcc -Wall test.c
$ ./a.out 
This program converts seconds to minutes and seconds. 
Just enter the number of seconds. 
Enter 0 to end the program. 
sec = 32767
But when I comment out the while statement:
#include<stdio.h>
#define S_TO_M 60
int main(void)
{
    int sec;
    //min,left;
    printf("This program converts seconds to minutes and ");
    printf("seconds. \n");
    printf("Just enter the number of seconds. \n");
    printf("Enter 0 to end the program. \n");
    printf("sec = %d\n",sec);
    /*
    while(sec > 0)
    {
        scanf("%d",&sec);
        min = sec/S_TO_M;
        left = sec % S_TO_M;
        printf("%d sec is %d min, %d sec. \n",sec,min,left);
        printf("Next input?\n");
    }
    */
    printf("Bye!\n");
    return 0;
}
Now GCC issues a warning and sec ends up being zero:
$ gcc -Wall test.c
test.c: In function ‘main’:
test.c:12:8: warning: ‘sec’ is used uninitialized in this function[-Wuninitialized]
printf("sec = %d\n",sec);
    ^
$ ./a.out
This program converts seconds to mintutes and seconds. 
Just enter the number of seconds. 
Enter 0 to end the program. 
sec = 0
Bye!
Why did the warning show up the second time but not the first time?
 
     
     
    