C Program to change the text colors.

Write a C Program to change the text colors.
In this program, we give the example of changing the text colors, background colors using conio.h library.
Syntax: void textcolor(int_color);
Here color is the integer variable, you can specify a color name also, but it should be a proper color name in capital letters.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

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* programs for commercial purposes,
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*
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#include<stdio.h>
#include<conio.h>
int main()
{
clrscr();
textcolor(BLUE); // Change font colour to blue
textbackground(WHITE); //change the background colour to white
cprintf("Color is Blue with white backgroundnn");
clrscr();
textcolor(RED+BLINK);//this one is blinking the text
cprintf("Color is RED with blinkingnn");
return 0;
}
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C program to sort a string

C Strings:
Write a C program to sort a string.
In this program we sort the string using bubble sort technique.
Bubble Sort is the simplest and easiest sorting technique. In this technique, the two successive items A[i] and A[i+1] are exchanged whenever A[i]>=A[i+1]. The larger values sink to the bottom of the array and hence it is called sinking sort. The end of each pass smaller values gradually “bubble” their way upward to the top(like air bubbles moving to surface of water) and hence called bubble sort.
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.

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* 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<string.h>
int main(){
    int i,j,n;
    char str[50],temp[50];
    printf("Enter a String:nn");
    scanf("%s",str);
    n=strlen(str);
    for(i=0;i<=n;i++)
    for(j=i+1;j<=n;j++){
        if(strcmp(str[i],str[j])>0){
            strcpy(temp,str[i]);
            strcpy(str[i],str[j]);
            strcpy(str[j],temp);
        }
    }
    printf("The sorted string is:%sn",str);
    return 0;
}
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C Aptitude Questions and answers with explanation

C Aptitude 7
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 7.1

  main()
{
printf("nab");
printf("bsi");
printf("rha");
}

Answer: hai

Explanation: n – newline b – backspace r – linefeed

C aptitude 7.2

  main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:45545

Explanation:The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the evaluation is from right to left, hence the result.

C aptitude 7.3

    #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:64

Explanation: the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

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C Aptitude Questions and answers with explanation

C Aptitude 6
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 6.1

   
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

Answer: SomeGarbageValue—1

Explanation: p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

C aptitude 6.2

 
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}


Answer:Compiler Error

Explanation:You should not initialize variables in declaration

C aptitude 6.3

      
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

Answer:Compiler Error

Explanation: The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

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C Aptitude Questions and answers with explanation.

C Aptitude 5
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 5.1

    #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}

Answer: sizeof(i)=1

Explanation: Since the #define replaces the string int by the macro char

C aptitude 5.2

main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}


Answer:i=0

Explanation:In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

C aptitude 5.3

     #include
main()
{
char s[]={'a','b','c','n','c',''};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:77

Explanation: p is pointing to character ‘n’. str1 is pointing to character ‘a’ ++*p. “p is pointing to ‘n’ and that is incremented by one.” the ASCII value of ‘n’ is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to ‘a’ that is incremented by 1 and it becomes ‘b’. ASCII value of ‘b’ is 98. Now performing (11 + 98 – 32), we get 77(“M”); So we get the output 77 :: “M” (Ascii is 77).

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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 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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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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