May 25, 2012

K & R C Chapter 7 Exercise Solutions.

We have already provided solutions to all the exercises in the book "C 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!='\0') ++*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 ‘\0’ 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("%d\n",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("%d\n",100);
}
 Note: 100; is an executable statement but with no action. So it doesn't give any problem

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May 24, 2012

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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May 23, 2012

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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May 21, 2012

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','\0'};
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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