# C Aptitude: Endianness, Pointer Arithmetic

C Aptitude 31

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) );}   `

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);}`

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);}`

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 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;}`

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);}`

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]();} `

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 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);}`

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

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);}`

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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# 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.

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact info@c-program-example.com
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* and browse!
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#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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Graphics in C
C Strings
C Aptitude```

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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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# 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;}`

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--);}`

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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# 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.

–>

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# C Aptitude Questions and answers with explanation

C Aptitude 26
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 26.1

` 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);}`

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 26.2

`   main ( ){ static char *s[ ]  = {“black”, “white”, “yellow”, “violet”}; char **ptr[ ] = {s+3, s+2, s+1, s}, ***p; p = ptr; **++p; printf(“%s”,*--*++p + 3);}`

Explanation:In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

C aptitude 26.3

`     main(){ int  i, n; char *x = “girl”; n = strlen(x); *x = x[n]; for(i=0; i < n; ++i)   {printf(“%sn”,x);x++;   } }`

(blank space)
irl
rl
l

Explanation: Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘’) to the first location. Now the string becomes “irl” . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

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# C program to hide the mouse pointers.

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.

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

1. Chapter 1: A Tutorial Introduction
2. Chapter 2: Types, Operators and Expressions
3. Chapter 3: Control Flow
4. Chapter 4: Functions and Program Structure
5. Chapter 5: Pointers and Arrays
6. Chapter 6: Structures
7. Chapter 7: Input and Output
You can purchase the book from here or here.