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C Aptitude 2
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The C aptitude questions and answers for those questions along with explanation for the interview related queries.

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Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

Predict the output or error(s) for the following:

C aptitude 2.1

 main()
            {
            static int var = 5;
            printf("%d ",var--);
            if(var)
                        main();
            }

Answer: 5 4 3 2 1

Explanation: When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

C aptitude 2.2

main()
{
             int c[ ]={2.8,3.4,4,6.7,5};
             int j,*p=c,*q=c;
             for(j=0;j<5;j++) {
                        printf(" %d ",*c);
                        ++q;     }
             for(j=0;j<5;j++){
printf(" %d ",*p);
++p;     }
}

Answer: 2 2 2 2 2 2 3 4 6 5

Explanation:Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

C aptitude 2.3

 main()
{
            extern int i;
            i=20;
printf("%d",i);
}

Answer:Linker Error : Undefined symbol ‘i’

Explanation: extern storage class in the following declaration, extern int i; specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

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#include <stdio.h>

static char daytab[2][13] = {
{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
};

/* day_of_year: set day of year from month & day */
int day_of_year(int year, int month, int day)
{
int i, leap;

if (year < 1752 || month < 1 || month > 12 || day < 1)
return -1;

leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;
if (day > daytab[leap][month])
return -1;

for (i = 1; i < month; i++)
day += daytab[leap][i];
return day;
}

/* month_day: set month, day from day of year */
int month_day(int year, int yearday, int *pmonth, int *pday)
{
int i, leap;

if (year < 1752 || yearday < 1)
return -1;

leap = (year%4 == 0 && year%100 != 0) || year%400 == 0;
if ((leap && yearday > 366) || (!leap && yearday > 365))
return -1;

for (i = 1; yearday > daytab[leap][i]; i++)
yearday -= daytab[leap][i];
*pmonth = i;
*pday = yearday;

return 0;
}

/* main: test day_of_year and month_day */
int main(void)
{
int year, month, day, yearday;

for (year = 1970; year <= 2000; ++year) {
for (yearday = 1; yearday < 366; ++yearday) {
if (month_day(year, yearday, &month, &day) == -1) {
printf("month_day failed: %d %dn",
year, yearday);
} else if (day_of_year(year, month, day) != yearday) {
printf("bad result: %d %dn", year, yearday);
printf("month = %d, day = %dn", month, day);
}
}
}

return 0;
}

C Program Examples

static

C Aptitude Questions and answers with explanation.

K & R C Programs Exercise 5-8.