C Aptitude Questions and answers with explanation

Posted on August 25, 2012June 19, 2017 by Sandeepa Nadahalli

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.

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 24.1

  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 24.2

   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 24.3

      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 program to hide the mouse pointers.

Posted on August 14, 2012June 19, 2017 by Sandeepa Nadahalli

C program to hide the mouse pointers.
In this program, we hide the mouse pointers using graphics.h and dos.h library functions.
To use graphics.h, we have to install the drivers in to the the system by using the initgraph() function.
In this program,we also check the condition that mouse pointer is available or not. 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<graphics.h>
#include<conio.h>
#include<dos.h>

void showmouseptr();
void hidemouseptr();

union REGS i, o;

int main()
{
    int  count = 1, gd = DETECT, gm;

    initgraph(&gd,&gm,"C:\TC\BGI");/*check your path, to install the drivers in to the the system*/

    i.x.ax = 0;
    int86(0X33,&i,&o);
    if(o.x.ax == 0)
    {
        printf("nt Sorry! Mouse support not available !");
    }
    else
    {
        showmouseptr();

        while(count<=10)
        {
            getch();
            count++;
            if(count%2==0)
            hidemouseptr();
            else
            showmouseptr();
        }
    }

    getch();
    return 0;
}


void showmouseptr()
{
    i.x.ax = 1;
    int86(0X33,&i,&o);
}

void hidemouseptr()
{
    i.x.ax = 2;             // to hide mouse
    int86(0X33,&i,&o);
}

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K & R C Exercise Solutions.

Posted on July 27, 2012June 19, 2017 by Sandeepa Nadahalli

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 chapters of the “C Programming Language (2nd Edition)” popularly known as K & R C book for easy reference.

Chapter 1: A Tutorial Introduction

Chapter 2: Types, Operators and Expressions

Chapter 3: Control Flow

Chapter 4: Functions and Program Structure

Chapter 5: Pointers and Arrays

Chapter 6: Structures

Chapter 7: Input and Output

You can purchase the book from here or here.

To get regular updates on new C programs, you can Follow @c_program

You can discuss these programs on our Facebook Page. Start a discussion right now,

our page!

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To browse more C Programs visit this link

C Aptitude Questions and answers with explanation

Posted on July 16, 2012June 19, 2017 by Sandeepa Nadahalli

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 ‘