C Aptitude Questions and answers with explanation

C Aptitude 24
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

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

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
Read more about how to prepare and write the C Aptitude Exams?
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 24.1

main()

{

char str[4]="test";

printf("%s",str);


}

Answer: test#$634278621@!

Explanation: The character array has the memory just enough to hold the string “test” and doesn’t have enough space to store the terminating null character. So it prints the test correctly and continues to print garbage values till it accidentally comes across a NULL character.

C aptitude 24.2

 
main()

{

char * str = "hello";

char * ptr = str;

char least = 127;

while (*ptr++)

least = (*ptr <least ) ?*ptr :least;

printf("%d",least);

}

Answer: 0

Explanation: After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

C aptitude 24.3

     main()

{

int i = 257;

int *iPtr = &i;

printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );

}

Answer: 11

Explanation: The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

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Write a C program to reverse a string using pointers.

C Strings:
Write a C program to reverse a string using pointers.
In this program, we reverse the given string by using the pointers. Here we use the two pointers to reverse the string, strptr holds the address of given string and in loop revptr holds the address of the reversed string.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
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* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
int main(){
int i=-1;
char str[100];
char rev[100];
char *strptr = str;
char *revptr = rev;
printf("Enter the string:n");
scanf("%s",str);
while(*strptr)
{
strptr++;
i++;
}
while(i >=0) {
strptr--;
*strptr = *revptr;
revptr++;
--i;
}
printf("nn Reversed string is:%s",rev);
return 0;
}



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C Aptitude Questions and answers with explanation

C Aptitude 19
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

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

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program.
Some of the illustrated examples will be from the various companies, and IT industry experts.
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 19.1

  main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}


Answer: 1==1 is TRUE

Explanation: When two strings are placed together (or separated by white-space) they are concatenated (this is called as “stringization” operation). So the string is as if it is given as “%d==1 is %s”. The conditional operator( ?: ) evaluates to “TRUE”.

C aptitude 19.2

  main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}



Answer: 2000 is a leap year

Explanation: An ordinary program to check if leap year or not.

C aptitude 19.3

     #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}

Answer: Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation: arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error. Rule of Thumb: #defines are used for textual replacement whereas typedefs are used for declaring new types.

C aptitude 19.4

      int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}

Answer: 30,20,10

Explanation: a'{‘ introduces new block and thus new scope. In the innermost block i is declared as, const volatile unsigned which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i’s value as 10.

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C Aptitude Questions and answers with explanation.

C Aptitude 4
C program is one of most popular programming language which is used for core level of coding across the board. C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

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

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program. Some of the illustrated examples will be from the various companies, and IT industry experts.
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 4.1

    main()
{
printf("%x",-1<<4);
}

Answer: fff0

Explanation: -1 is internally represented as all 1’s. When left shifted four times the least significant 4 bits are filled with 0’s.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

C aptitude 4.2

main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}


Answer:Compiler Error : Type mismatch in redeclaration of function display

Explanation:In third line, when the function display is encountered, the compiler doesn’t know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

C aptitude 4.3

   main()
{
int c=- -2;
printf("c=%d",c);
}

Answer:c=2;

Explanation: Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus. Note: However you cannot give like –2. Because — operator can only be applied to variables as a decrement operator (eg., i–). 2 is a constant and not a variable.

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C Aptitude Questions and answers with explanation.

C Aptitude 3
C program is one of most popular programming language which used for core level of coding across the board. Now a days C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

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

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program. Some of the illustrated examples will be from the various companies, and IT industry experts.
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 3.1

   main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

Answer: 0 0 1 3 1

Explanation:Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

C aptitude 3.2

main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}


Answer:12

Explanation:The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

C aptitude 3.3

  main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}



Answer:three

Explanation: The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn’t match.

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C Aptitude Questions and answers with explanation.

C Aptitude 2
C program is one of most popular programming language which used for core level of coding across the board. Now a days C program is used for operating systems, embedded systems, system engineering, word processors,hard ware drivers, etc.

In this site, we have discussed various type of C programs till date and from now on, we will move further by looking at the C aptitude questions.

In the coming days, we will post C aptitude questions, answers and explanation for interview preparations.

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

We hope that this questions and answers series on C program will help students and freshers who are looking for a job, and also programmers who are looking to update their aptitude on C program. Some of the illustrated examples will be from the various companies, and IT industry experts.
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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C Program to convert number to words.

Problem Statement

Write a C Program to convert number to words.
Example:
Input: 4562
Output: four thousand five hundred sixty two

Solution

The program takes a number between 0 and 99999 as input. First, we count the total number of digits in the number. This is done by successively dividing the number by 10 until we reach 0. Next we go through each digit in the number and decide what to print based on the value of digit at that position. Digit at a particular position can be obtained by dividing the number by a divider. Divider for a given position is pow(10, (position-1)). We start from the left most position and move right as we identify digit at each position and print relevant words to screen.

In the above example, we first divide the number(4562) by 1000 and get 4. We print four thousand and move on to next digit. The next number to work with can be obtained by num % divider. In this case 4562 % 1000 = 562. We divide this number by 100 (divider for position 3) to get 5. So, we print five hundred and move on to next digit. So forth and so on.

Special case handling

A value of 1 at position 2 and 5 needs special attention. Take these examples. In 23, when we encounter 2 at position 2, we can be print ‘twenty’ without knowing next digit. It will be taken care of next. But in case of 15, when we encounter 1 in position 2, we can’t print anything yet! It could be any number between eleven and nineteen. So, we have to wait till we see the next digit. To keep track of this, we maintain a flag variable which is set to 1 to indicate such a case.

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

The Program

#include<stdlib.h>
#include<stdio.h>
void main() {
long num, div, n1;
int flag, digit, pos, tot_dig;
printf("\nEnter a number: ");
scanf("%ld", &num);
if(num == 0) {
printf("Zeron\n");
exit(0);
}
if(num > 99999) {
printf("please enter a number between 0 and 100000\n\n");
exit(0);
}
tot_dig = 0;
div = 1;
n1 = num;
while ( n1 > 9 ) {
n1 = n1 / 10;
div = div * 10;
tot_dig++;
}
tot_dig++;
pos = tot_dig;
while ( num != 0 ) {
digit = num / div;
num = num % div;
div = div / 10;
switch(pos) {
case 2:
case 5:
if ( digit == 1 )
flag = 1;
else {
flag = 0;
switch(digit) {
case 2: printf("twenty ");break;
case 3: printf("thirty ");break;
case 4: printf("forty ");break;
case 5: printf("fifty ");break;
case 6: printf("sixty ");break;
case 7: printf("seventy ");break;
case 8: printf("eighty ");break;
case 9: printf("ninty ");
}
}
break;
case 1:
case 4:
if (flag == 1) {
flag = 0;
switch(digit) {
case 0 : printf("ten ");break;
case 1 : printf("eleven ");break;
case 2 : printf("twelve ");break;
case 3 : printf("thirteen ");break;
case 4 : printf("fourteen ");break;
case 5 : printf("fifteen ");break;
case 6 : printf("sixteen ");break;
case 7 : printf("seventeen ");break;
case 8 : printf("eighteen ");break;
case 9 : printf("nineteen ");
}
} else {
switch(digit) {
case 1 : printf("one ");break;
case 2 : printf("two ");break;
case 3 : printf("three ");break;
case 4 : printf("four ");break;
case 5 : printf("five ");break;
case 6 : printf("six ");break;
case 7 : printf("seven ");break;
case 8 : printf("eight ");break;
case 9 : printf("nine ");
}
}
if (pos == 4)
printf("thousand ");
break;
case 3:
if (digit > 0) {
switch(digit) {
case 1 : printf("one ");break;
case 2 : printf("two ");break;
case 3 : printf("three ");break;
case 4 : printf("four ");break;
case 5 : printf("five ");break;
case 6 : printf("six ");break;
case 7 : printf("seven ");break;
case 8 : printf("eight ");break;
case 9 : printf("nine ");
}
printf("hundred ");
}
break;
}
pos--;
}
if (pos == 4 && flag == 0)
printf("thousand");
else if (pos == 4 && flag == 1)
printf("ten thousand");
if (pos == 1 && flag == 1)
printf("ten ");
}
view raw number_to_words.c hosted with ❤ by GitHub

 

Sample Output

Sample output

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C program to merge two arrays.

Arrays in C.
Write a C program to merge two arrays.
In this program we check the elements of arrays A, B and put that elements in the resulted array C in sorted manner.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
#include<conio.h>
void main( )
{
int n,m,i,j,k,c[40],a[20],b[20];
clrscr ();
printf("Enter how many elements for array A?:n");
scanf("%d",&n);
printf ("Enter how many elements for array B?:n");
scanf("%d",&m);
printf("Enter elements for A:-n");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
printf("Enter elements for B:-n");
for(j=0;j<m;j++)
scanf("%d",&b[j]);
i=j=k=0;
while(i<n&&j<m)
{
if(a[i]<b[j])
c[k++]=a[i++];
else
if(a[i]>b[j])
c[k++]=b[j++];
else

{
c[k++]=b[j++];
i++;
j++;
}
}

if(i<n)
{
int t;
for(t=0;t<n;t++)

c[k++]=a[i++];
}
if(j<m)
{
int t;
for(t=0;t<m;t++)
{
c[k++]=b[j++];
}
}
printf("nn Merged Array C:nn")
for(k=0;k<(m+n);k++)
printf("t n %d ",c[k]);
getch();
}
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C Program to find day of birth from DOB

Problem Statement

Write a C Program to find the day of birth when date of birth is given. For example, if we give 02/04/2107  (02nd April 2017) as input, the program should output that it was a Sunday.

Solution

  1. Pick a year as base year where January 1st falls on a Monday. We are using 1900 as base year in our program.
  2. Then we find the number of years after base year. Day of the week in the beginning of the year shifts by 1 in every non leap year. 365 days -> 52 * 7 + 1. 
  3. Once we know on what day a given year started, we can calculate each month’s starting day.
  4. From there it’s as simple as counting the days and finding which day it is.

In this program we find the day of birth like Monday, Sunday.. when date of birth is given. For a more detailed explanation of the method can be found here Converting a Month-Date-Year to the Day of the Week

The Program

#include<stdio.h>
#include<stdlib.h>
int main() {
int d, m, y, year, month, day, i;
printf("Enter date of birth (DD MM YYYY) :");
scanf("%d %d %d", &d, &m, &y);
if( (d > 31) || (m > 12) || (y < 1900 || y >= 2100) )
{
printf("INVALID INPUT. Please enter a valid date between 1900 and 2100");
exit(0);
}
year = y-1900;
year = year/4;
year = year+y-1900;
switch(m)
{
case 1:
case 10:
month = 1;
break;
case 2:
case 3:
case 11:
month = 4;
break;
case 7:
case 4:
month = 0;
break;
case 5:
month = 2;
break;
case 6:
month = 5;
break;
case 8:
month = 3;
break;
case 9:
case 12:
month = 6;
break;
}
year = year + month;
year = year + d;
/* Need to make sure extra day is not needed in leap year for dates before March */
if(( y > 1900 ) && ( y % 4 == 0 ) && ( m < 2 ) )
year--;
day = year % 7;
switch(day)
{
case 0:
printf("Day is SATURDAY\n");
break;
case 1:
printf("Day is SUNDAY\n");
break;
case 2:
printf("Day is MONDAY\n");
break;
case 3:
printf("Day is TUESDAY\n");
break;
case 4:
printf("Day is WEDNESDAY\n");
break;
case 5:
printf("Day is THURSDAY\n");
break;
case 6:
printf("Day is FRIDAY\n");
break;
}
return 0;
}
view raw day_of_birth.c hosted with ❤ by GitHub

Sample Output

Date of birth to day of birth
Finding day of birth from DOB

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C program to print all the possible permutations of given digits

Write a C program to print all the possible permutations of given digits.
Permutations means possible way of rearranging in the group or set in the particular order.
Example:
Input:1, 2, 3
Output:1 2 3, 1 3 2, 2 1 3, 3 1 2, 2 3 1, 3 2 1
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
int lev=-1,n,val[50],a[50];
void main()
{
int i,j;
clrscr();
printf("Enter how many numbers?n");
scanf("%d",&n);
printf("nEnter %d numbers:nn",n);
for(i=0;i<n;i++)
{
val[i]=0;
j=i+1;
scanf("%dnn",&a[j]);
}
visit(0);
getch();
}
visit(int k)
{
int i;
val[k]=++lev;
if(lev==n)
{
for(i=0;i<n;i++)
printf("%2d",a[val[i]]);
printf(" ");
}
for(i=0;i<n;i++)
if(val[i]==0)
visit(i);
lev--;
val[k]=0;

}
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