K & R C Chapter 7 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 7 of the book for easy reference.

Chapter 7: Input and Output

  1. Exercise 7-1. Write a program that converts upper case to lower or lower case to upper, depending on the name it is invoked with, as found in argv[0].
    Solution to Exercise 7-1.
  2. Exercise 7-2.Write a program that will print arbitrary input in a sensible way. As a minimum, it should print non-graphic characters in octal or hexadecimal according to local custom, and break long text lines.
    Solution to Exercise 7-2.
  3. Exercise 7-3.Revise minprintf to handle more of the facilities of printf .
    Solution to Exercise 7-3.
  4. Exercise 7-4. Write a private version of scanf analogous to minprintf from the previous section.
    Solution to Exercise 7-4.
  5. Exercise 7-5. Rewrite the postfix calculator of Chapter 4 to use scanf and/or sscanf to do the input and number conversion.
    Solution to Exercise 7-5.
  6. Exercise 7-6. Write a program to compare two files, printing the first line where they differ.
    Solution to Exercise 7-6.
  7. Exercise 7-7. Modify the pattern finding program of Chapter 5 to take its input from a set of named files or, if no files are named as arguments, from the standard input. Should the file name be printed when a matching line is found?
    Solution to Exercise 7-7.
  8. Exercise 7-8. Write a program to print a set of files, starting each new one on a new page, with a title and a running page count for each file.
    Solution to Exercise 7-8.
  9. Exercise 7-9. Functions like isupper can be implemented to save space or to save time. Explore both possibilities.
    Solution to Exercise 7-9.
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C Aptitude Questions and answers with explanation

C Aptitude 8
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 8.1

  main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='') ++*p++;
printf("%s %s",p,p1);
}


Answer:ibj!gsjfoet

Explanation: ++*p++ will be parse in the given order Ø *p that is value at the location currently pointed by p will be taken Ø ++*p the retrieved value will be incremented Ø when ; is encountered the location will be incremented that is p++ will be executed Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘’ and p1 points to p thus p1doesnot print anything.

C aptitude 8.2

  #define a 10
main()
{
#define a 50
printf("%d",a);
}

Answer:50

Explanation:The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

C aptitude 8.3

    #define clrscr() 100
main()
{
clrscr();
printf("%dn",clrscr());
}

Answer:100

Explanation: Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
 printf(“%dn”,100);
}
 Note: 100; is an executable statement but with no action. So it doesn’t give any problem

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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 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 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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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 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.

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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<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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