May 18, 2012

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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May 17, 2012

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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May 16, 2012

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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May 15, 2012

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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May 14, 2012

C Program to convert number to words.

Write a C Program to convert number to words.
Example:
Input: 4562
Output: four thousand five hundred sixty two
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 < conio.h >
void main()
{
    long num,div,n1;
    int flag,digit,pos,tot_dig;
    
    clrscr();
    printf("\n Enter number: ");
    scanf("%ld",& num);
    if(num==0)
    {
        printf("Zero\n\n");
        exit(0);
    }
    if(num>99999)
    {
        printf("please enter 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("ninteen ");
                }
            }
            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 ");
    getch();
}
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