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C Aptitude: Endianness, Pointer Arithmetic

C Aptitude 31
In this episode we’ll learn learn more about Endianness, Pointer Arithmetic.

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 31.1

  
main() {
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}

Answer: 2 1

Explanation: The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the INTEL machines are ‘small-endian’ machines. Small-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. The integer value 258 is stored in memory as: 00000001 00000010.

C aptitude 31.2

main() {
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}

Answer:556

Explanation:The integer value 300  in binary notation is: 00000001 00101100. It is  stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it  is  00000010 00101100 => 556.

C aptitude 31.3

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 ‘’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

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C Program to implement bucket sort

Write C program to implement bucket sort.
The idea of Bucket Sort is to divide the interval [0, 1] into n equal-sized sub intervals, or buckets, and then distribute the n input numbers into the buckets. To produce the output, we simply sort the numbers in each bucket and then go through the buckets in order, listing elements in each.
Bucket sort runs in linear time when the input is drawn from a uniform distribution. Read more about C Programming Language .
/***********************************************************
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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
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* and browse!
* 
*                      Happy Coding
***********************************************************/

#include<stdio.h>
void Bucket_Sort(int array[], int n)
{   
 int i, j;   
 int count[n];  
 for(i=0; i < n; i++)
 {   
  count[i] = 0;   
 }     
 for(i=0; i < n; i++)
 {    
  (count[array[i]])++; 
 }     
 for(i=0,j=0; i < n; i++)
 {   
  for(; count[i]>0;(count[i])--) 
  {       
   array[j++] = i; 
  }  
 }   
}    
int main() 
{ 
 int array[100];   
 int num;   
 int i;  
 printf("Enter How many Numbers : ");    
 scanf("%d",&num);    
 printf("Enter the %d elements to be sorted:n",num);  
 for(i = 0; i < num; i++ )
 {   
  scanf("%d",&array[i]);  
 }   
 printf("nThe array of elements before sorting : n"); 
 for (i = 0;i < num;i++) 
 {    
  printf("%d ", array[i]);   
 }    
 printf("nThe array of elements after sorting : n");  
 Bucket_Sort(array, num);  
 for (i = 0;i < n;i++) 
 {     
  printf("%d ", array[i]);  
 }   
 printf("n");      
 return 0; 
} 

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C Sorting

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C Aptitude Questions and answers with explanation

C Aptitude 30
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 30.1

  main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}

Answer:1000

Explanation: Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

C aptitude 30.2

   main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}

Answer: g20fy

Explanation: Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

C aptitude 30.3

     # include<stdio.h>
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}

Answer: bye

Explanation: ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

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C Aptitude Questions and answers with explanation

C Aptitude 29
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 29.1

  enum colors {BLACK,BLUE,GREEN}
main()
{

printf("%d..%d..%d",BLACK,BLUE,GREEN);

return(1);
}

Answer: 0..1..2

Explanation: enum assigns numbers starting from 0, if not explicitly defined.

C aptitude 29.2

   void main()
{
char far *farther,*farthest;

printf("%d..%d",sizeof(farther),sizeof(farthest));

}


Answer: 4..2

Explanation: The second pointer is of char type and not a far pointer

C aptitude 29.3

     main()
{
int i=400,j=300;
printf("%d..%d");
}

Answer: 400..300

Explanation: printf takes the values of the first two assignments of the program. Any number of printf’s may be given. All of them take only the first two.

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C Aptitude Questions and answers with explanation

C Aptitude 28
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.

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 28.1

 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
 *p that is value at the location currently pointed by p will be taken
  ++*p the retrieved value will be incremented
 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 28.2

   #include <stdio.h>
#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 28.3

     
main()
{
printf("%p",main);
}


Answer:Some address will be printed.
Explanation: Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

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–>
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C Program to delete a file using remove function.

Write a C program to delete a specified file using remove function.
remove function deletes the specified file, where it takes file name as the argument and if file is successively deleted it returns 0 , other wise returns a non zero value.
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 [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
* 
*                      Happy Coding
***********************************************************/

#include<stdio.h>
#include<string.h>
int main()
{
    char file_name[40];
    printf("Enter the file name with location:n Example, c:example.txtnn");
    gets(file_name);
    if(remove(file_name) != 0 )
    puts("Error deleting filenn");
    else
    puts("File successfully deletednn");
    return 0;
}
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C Strings
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C program without using main function

Write a C program without using main function.
In this program, we didn’t use the main function, but we are using the preprocessor directive #define with arguments to define the Main function.
A Preprocessor is program which processes the source code before compilation.
The ‘##‘ operator is called the token pasting or token merging operator. In the second line of program we used it to merges the 4th,1st,3rd & the 2nd characters(tokens), and it compiles as “msut”.
When the controller calls int begin, preprocessor replaces the macro “begin” with the expansion decode(a,n,i,m,a,t,e). As defined in the macro decode it merges the 4th,1st,3rd & the 2nd characters, and int begin becomes int main(), so program runs successfully. Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.
–>

/***********************************************************
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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>
#define decode(s,t,u,m,p,e,d) m##s##u##t
#define begin decode(a,n,i,m,a,t,e)

int begin()

{
printf(” Ha HA see how it is?? “);
}
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C Aptitude Questions and answers with explanation

C Aptitude 27
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.

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 27.1

 main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}

Answer: 1000

Explanation: Normally the return value from the function is through the information from the accumulator. Here _AX is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

C aptitude 27.2
// If the inputs are 0,1,2,3 find the o/p

  int i;
main()
{
int t;
for ( t=4;scanf("%d",&i)-t;printf("%dn",i))
printf("%d--",t--);
}

Answer: 4–0 3–1 2–2

Explanation:Let us assume some x= scanf(“%d”,&i)-t the values during execution will be, t i x 4 0 -4 3 1 -2 2 2 0

C aptitude 27.3

     
int a,b;
func(a,b)

{
return( a= (a==b) );
}
main()
{
int process(),func();
printf("The value of process is %d !n ",process(func,3,6));
}
int val1,val2;
process(pf,val1,val2)
int (*pf) ();
{
return((*pf) (val1,val2));
}

Answer:The value if process is 0 !
Explanation: The function ‘process’ has 3 parameters – 1, a pointer to another function2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function ‘func’. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function ‘process’.

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–>
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C program to print given numbers as a pyramid

Write a C program to print given numbers as a pyramid.
In this program we use the simple for statements to produce the patterns.
This program prints the following output,
        1
      232
    34543
  4567654
567898765
 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 [email protected]
* To find more C programs, do visit www.c-program-example.com
* and browse!
*
* Happy Coding
***********************************************************/

#include<stdio.h>


void display(int start, char *str){

int i,j;


for(i=start; i<=(start+start); i++)

printf("%c",str[i]);

for(j=i-2;j>=start;j--)

printf("%c",str[j]);

}


main(){

char str[]="123456789";

int i, j;

for(i=0;i<5;i++){

display(i, str);

printf("n");

}

}
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Graphics in C

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