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 Aptitude Questions and answers with explanation.

C Aptitude 1
C program aptitude questions, answers and explanation for interview preparations.
In this site, We discussed various type of C programs, now we are moving into further steps by looking at the c aptitude questions.
This questions and answers are helpful to freshers and job hunters. C interview questions are from various companies and 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 1.1

void main()
{
int const * p=5;
printf("%d",++(*p));
}

Answer: Compiler error: Cannot modify a constant value.

Explanation: p is a pointer to a “constant integer”. But we tried to change the value of the “constant integer”.

C aptitude 1.2

main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}


Answer: mmmm aaaa nnnn

Explanation: s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

C aptitude 1.3

 main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}


Answer: I hate U

Explanation: For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double. Rule of Thumb: Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

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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 [email protected]
* 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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K & R C Chapter 6 Exercise Solutions.

We have already provided solutions to all the exercises in the bookC Programming Language (2nd Edition) popularly known as K & R C book.

In this blog post I will give links to all the exercises from Chapter 6 of the book for easy reference.

Chapter 6: Structures

  1. Exercise 6-1. Our version of getword does not properly handle underscores, string constants, comments, or preprocessor control lines. Write a better version.
    Solution to Exercise 6-1.
  2. Exercise 6-2.Write a program that reads a C program and prints in alphabetical order each group of variable names that are identical in the first 6 characters but different somewhere thereafter. Don’t count words within strings and comments. Make 6 a parameter that can be set from the command line.
    Solution to Exercise 6-2.
  3. Exercise 6-3.Write a cross-referencer that prints a list of all words in a document, and, for each word, a list of the line numbers on which it occurs. Remove noise words like “the,” “and,” and so on.
    Solution to Exercise 6-3.
  4. Exercise 6-4. Write a program that prints the distinct words in its input sorted into decreasing order of frequency of occurrence. Precede each word by its count.
    Solution to Exercise 6-4.
  5. Exercise 6-5. Write a function undef that will remove a name and definition from the table maintained by lookup and install .
    Solution to Exercise 6-5.
  6. Exercise 6-6. Implement a simple version of the #define processor (i.e., no arguments) suitable for use with C programs, based on the routines of this section. You may also find getch and ungetch helpful.
    Solution to Exercise 6-6.
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C Program to search the linked list.

Data structures using C,
Write c program to search the linked list.
Linked list is a data structure in which the objects are arranged in a linear order. In this program, we sort the list elements in ascending order. 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 [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
struct l_list
{
int info;
struct link *next;
}
start, *node;

int search(int);
void main()
{
int no,i,item,pos;
clrscr();
start.next=NULL;
node=&start;
printf("How many nodes, you want in linked list? ");
scanf("%d",&no);
printf("");
for(i=0;i<no;i++)
{
node->next=(struct l_list *)malloc(sizeof(struct l_list));
printf("Enter element in node %d: ",i+1);
scanf("%d",&node->info);
node=node->next;
}
node->next=NULL;
printf("Linked list(only with info field) is:");

node=&start;
while(node->next!=NULL)
{
printf("%d ",node->info);
node=node->next;
}
printf("

Enter item to be searched : ");
scanf("%d",&item);
pos=search(item);
if(pos<=no)
printf("Your item is at node %d",pos);
else
printf("Sorry! item is no in linked list.a");
getch();
}

int search(int item)
{
int n=1;
node=&start;
while(node->next!=NULL)
{
if(node->info==item)
break;
else
n++;
node=node->next;
}
return n;
}
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C program to demonstrate assert macro.

Write a C program to demonstrate assert macro.
assert macro defined in assert.h library. assert is mainly used for debugging. assert function checks the conditions at run time, and program executes only if the assert is true otherwise program aborts and shows the error message.
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 [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include <stdio.h>
#include <assert.h>
#include<conio.h>
int main()
{
int num1, num2;
clrscr();
printf("Enter two numbers:nn");
scanf("%d %d", &num1, &num2);
assert(num2 != 0);
printf("%d/%d = %.2fn", num1, num2, num1/(float)num2);
return 0;
}
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C program to implement SJF algorithm.

Write a C program to implement SJF algorithm.
Shortest Job First(SJF) is the CPU scheduling algorithm. In SJF, if the CPU is available it allocates the process which has smallest burst time, if the two process are same burst time it uses FCFS algorithm.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 [email protected]
* 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 i,j,n,brust_time[10],start_time[10],end_time[10],wait_time[10],temp,tot;
float avg;
clrscr();
printf("Enter the No. of jobs:nn");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
printf("n n Enter %d process burst time:n",i);
scanf("%d",&brust_time[i]);
}

for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
if(brust_time[i]>brust_time[j])
{
temp=brust_time[i];
brust_time[i]=brust_time[j];
brust_time[j]=temp;
}
}

if(i==1)
{
start_time[1]=0;
end_time[1]=brust_time[1];
wait_time[1]=0;
}

else
{
start_time[i]=end_time[i-1];
end_time[i]=start_time[i]+brust_time[i];
wait_time[i]=start_time[i];
}
}
printf("nn BURST TIME t STARTING TIME t END TIME t WAIT TIMEn");
printf("n ********************************************************n");
for(i=1;i<=n;i++)
{
printf("n %5d %15d %15d %15d",brust_time[i],start_time[i],end_time[i],wait_time[i]);
}
printf("n ********************************************************n");
for(i=1,tot=0;i<=n;i++)
tot+=wait_time[i];
avg=(float)tot/n;
printf("nnn AVERAGE WAITING TIME=%f",avg);
for(i=1,tot=0;i<=n;i++)
tot+=end_time[i];
avg=(float)tot/n;
printf("nn AVERAGE TURNAROUND TIME=%f",avg);
for(i=1,tot=0;i<=n;i++)
tot+=start_time[i];
avg=(float)tot/n;
printf("nn AVERAGE RESPONSE TIME=%fnn",avg);
getch();
}
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