Multiple Data Compression and Encryption using Iterative Technique. B97

spatlou

Minutes of Progress

saurabh.hmr009

Can anyone tell whats wrong in this program of binary search?

#include<stdio.h>
#include<stdlib.h>
int main()
{
   int *arr;
   int i,mid,p,low=0,high=9,num;
   arr=(int *)malloc(sizeof(int)*10);
   if(!arr)
   {
       perror("No pointer");
       exit(EXIT_FAILURE);
   }
   printf("Enter the elements of array: \n");
   for(i=0;i<10;i++)
       scanf("%d",arr+i);

   printf("enter the number to be searched: \n");
   scanf("%d",&num);


   while(high>=low)
   {
       mid=low+high/2;
       p=mid;

       if(*(arr+p)==num)
       {   printf("Number found at indeex number: %d \n",mid);
           break;
       }
       else if(*(arr+p)>num)
           high=p-1;
       else if (*(arr+p)<num)
           low=p+1;
       else
           printf("Number doesnt exist in array\n");

   }
return 0;

}

vishwajeet.gaur1989

The program is correct however please ensure that you use ascending order sorted array only for the above logic. Also mid=low+high/2 is not correct way to calculate mid value (see precedence operator).

Please use mid=(low+high)/2.

saurabh.hmr009

yep saw that .

vishwajeet.gaur1989

int binary_search(int *ptr,int k)

{

int low,mid,high;

low=0;

high=9;

mid=(low+high)/2;

while(high>=low)

{

if(*(ptr+mid)==k)

return mid;

else if(*(ptr+mid)>k)

high=mid-1;

else if(*(ptr+mid)<k)

low=mid+1;

mid=(low+high)/2;

}

return 0;

}

sankalpnegi2310

In case of binary search, the basic prerequisite is to have a sorted array...so we can ask the user to input the same or..make another function in order to sort the random elements (moreover this should be the preferred logic) in the required order..ascending or descending ..

for eg.

int sorted(int *)

we can call this function inside our main(as mentioned herein)..we will now make our local pointer point to the value returned by the function sorted()..we can now perform binary search on this array..Thats it..rest of the code remains
the same..

adeshagarwal02

please check my program for insertion sort
this code is working but i want different way
#include<stdio.h>
#include<stdlib.h>
int *getArray();
int *insertion(int *);
int main()
{
    int i;
    int *ptr;
//    ptr =(int *)malloc(sizeof(int)*10);
    printf("Please enter the elements of 10 element array\n");
    ptr=getArray();
    printf("Insertion shot\n");
    ptr= insertion(ptr);
    for(i=0;i<10;i++)
    {
        printf("%d\n",*(ptr+i));
    }

    return 0;
}
int *getArray(void)
{
    int i;
    int *ptr;
    ptr=(int *)malloc(sizeof(int)*10);
    for(i=0;i<10;i++)
    {
        scanf("%d",ptr+i);

    }
    return ptr;
}

int *insertion(int *ptr)
{
    int i,j,s,t;
    for(i=0;i<10;i++)
    {
        s=i+1;
        j=0;

    for(j=0;j<=s;j++)
    {
        if(*(ptr+s)<*(ptr+j))
        {
            t=*(ptr+j);
            *(ptr+j)=*(ptr+s);
            *(ptr+s)=t;
        }
    }
}

    return ptr;
}

adeshagarwal02

And i have problem in array average
please check
whenever I input 'm' I want to break the loop but this array is not taking m
can anybody help to solve this so I can give input as an charterer in an integer array

#include<stdio.h>
int main()
{
    int arry[]={'\0'};
    int i,j=0,k,sum=0;
    float avg;
    char a;
    printf("please enter the value for average\n");
    printf("And enter m for average when all values are given\n");
    while(1)
    {
        scanf("%d",&arry[j]);
        printf("\n%d", j);
        a=arry[j];
        if(a=='m')
        {

            break;

        }

        j=j+1;
    }

    printf("Average is=");
    for(k=0;k<j;k++)
    {
        sum=sum+arry[k];
    }
    avg=(float)sum/j;
    printf("%f\n",avg);
    return 0;
}

sankalpnegi2310

scanf("%d",&arry[j]);
        printf("\n%d", j);
        a=arry[j];
        if(a=='m')

here a character array takes input as an int...so a=arry[j] will not be stored as a character ..therefore cannot be compared with 'm' which is a character..

however there may be other flaws due to conflicting concepts..

my suggestion:
use an array of pointers that is string type..then use atoi() function for int conversion
and proceed further..it will be much more conceptual coz here u have used character array and value dtored at a character will malfunction for more than 1-digit numbers..eg 123

adeshagarwal02

@sankalpnegi2310
okay
i will try and update

sankalpnegi2310

@adeshagarwal02 great

vishwajeet.gaur1989

atoi command can be used here however it has a drawback. If you enter string as 2abcd it will still take value 2 as an integer after conversion of this string. Following program illustrates the use of atoi with the same vulnerability.

#include<stdio.h>

int main()

{

char a[10];

int i=0,s=0;

while(1)

{

printf("ENTER INPUT VALUES \n");

scanf("%s",&a);

if(strcmp(a,"m")==0)

break;

s=s+atoi(a);

i++;

}

printf("AVERAGE OF %d NUMBER IS %d",i,s/i);

}

vishwajeet.gaur1989

@adeshagarwal02

Try using following for insertion sort-->

int *get_Array()

{

int i;

int *ptr;

ptr=(int *)malloc(sizeof(int)*10);

printf("ENTER 10 ELEMENTS TO SORT USING SELECTION SORT TECHNIQUE");

for(i=0;i<10;i++)

{

scanf("%d",ptr+i);

}

return ptr;

}

int insertion_sort(int *ptr)

{

int i,s,key;

for(i=1;i<10;i++)

{

key=*(ptr+i);

s=i-1;

while(s>=0 && key<*(ptr+s))

{

*(ptr+s+1)=*(ptr+s);

s--;

}

*(ptr+s+1)=key;

}

adeshagarwal02

@vishwajeet.gaur1989
okay
I will try this :)

adeshagarwal02

how to give path to a variable in shell script
#! /bin/bash
file= /home/adesh/emblogic/states
for state in `cat $file`
do
echo "Visit beautiful $state"
done

Please tell me ?

adeshagarwal02

why this is not working ?
#! /bin/bash
file= /home/adesh/emblogic/*
for files in $file
do
if [ -d $files ]
then
echo "$files it is a directory"
else
echo "$files is a file"
fi
done
please tell ?

vishwajeet.gaur1989

@adeshagarwal02

Try using VAR_NAME='/home/Vishwajeet/Documents/ShellScripts/Test'

sankalpnegi2310

plz post the exact error in order to debug @adeshagarwal02

just remove the space before '/' rest is cool...even without ' ' it should work as just a string

adeshagarwal02

@vishwajeet.gaur1989
Awesome

adeshagarwal02

@sankalpnegi2310
problem solved :)

vishwajeet.gaur1989

@adeshagarwal02

Single quotes could be avoided but please ensure you do not use spaces while assigning path to a variable.

Incorrect:: VAR_NAME ='/home/Vishwajeet/Documents/ShellScripts/Test'

Correct:: VAR_NAME='/home/Vishwajeet/Documents/ShellScripts/Test'

sankalpnegi2310

@adeshagarwal02
..for your secomd issue..use

file= /home/adesh/emblogic/ #remove *
for files in `ls $file1` #ls $file is within backticks

as just mentioning the path wont give access to file names...we need list (ls) cmd..
try this and let me know..

saurabh.hmr009

try like this
file='/home/saurabh/C_programs/Arrays/states'
with no space

saurabh.hmr009

@adeshagarwal02, I don't know if you want to check whether if its directory or a file.

like @sankalpnegi2310 mentioned , if you want to show files in a path you need to mention 'ls" command , like 'ls $file' in the program.

If you just use
for files in $files
it returns that emblogic is a directory.

if you use ls there then it will show emblogic as directory and files which that directory contains.

sankalpnegi2310

@saurabh.hmr009 @adeshagarwal02

theres a problem in my above concept as well..pls take note...this is imp

'ls' cmd wd give all the names but while comparisons we wd be comparing only with the names , when however we require exact file location..therefore it needs to be edited..as below:

#! /bin/bash
file=/home/xenial-xerus/      #this is my loaction to check..we can also use 'pwd' cmd as required

for i in `ls $file`
do

if [ -d $file/$i ]                        #here $file/$i =/home/xenial-xerus/$i=required location as we need full adress
then
        echo "$i it is a directory"
else
        echo "$i is a file"
fi
done

saurabh.hmr009

and try using just .... if [ $file ]

sankalpnegi2310

@saurabh.hmr009 @adeshagarwal02 wishes to check if its a file or directory..so he wd be needing more than just ..if [ $file ]

he wd need -f -d type of options to know that ...

however for syntax purposes one can also use

if test -d <filepath>

sankalpnegi2310

As a matter of fact just check out my 2 new scripts for system administration on the other forum named Batch 97 : Work log/Useful info

saurabh.hmr009

@sankalpnegi2310, yep got it

sankalpnegi2310

great then @saurabh.hmr009

adeshagarwal02

thank you both
@saurabh.hmr009 and @sankalpnegi2310
but i had already done :)
when wrote problem solved :)
please check it is working

#! /bin/bash
file='/home/adesh/emblogic/*'
for files in $file
do
if [ -d $files ]
then
echo "$files it is a directory"
else
echo "$files is a file"
fi
done

saurabh.hmr009

this one has same working too.

#! /bin/bash
file=/home/saurabh/C_programs/Array/
for files in `ls $file`
do
if [ -f $files ]
then
echo "$files it is a directory"
else
echo "$files is a file"
fi
done

saurabh.hmr009

Here Im trying to write the encrypted codes in the compressed file but the thing is it isnt allowing me to write anything in the Compressed file. If I try to write something else in the file the it is allowing me to write but when in loop, it isnt writing anything in file.

int compression4(int fd,char *ma)
{
   char ch,by,c;
   int i1,cfd,efd,ret;
   printf("\nStart --> compression\n");
   cfd=open("Compressed4",O_WRONLY|O_CREAT);
   if(cfd==-1)
   {   perror("Open");
       exit(EXIT_FAILURE);
   }

   while(read(fd,&ch,1))
   {
       by ^=by;
       i1=findindex(ch,ma);
       sprintf(&c,"%d",i1);
       c=c<<4;
       by|=c;
       read(fd,&ch,1);
       i1=findindex(ch,ma);
       sprintf(&c,"%d",i1);
       c=c<<4;
       c=c>>4;
       by|=c;
       write(cfd,&by,1);
       if(ret==-1);
       {
           perror("write() failed");
           exit(EXIT_FAILURE);
       }
   }

// write(efd,ma,strlen(ma));
   printf("\nContents of Encryption Key are: %s\n",ma);
   close(cfd);
}

saurabh.hmr009

in output it shows that perror which I declared in last ,"Write () failed" with some garbage

lkshchoudhary004

1) why we need return type in programming, why we need to hold a value in return type
2) what the mean of value in return type(like 0, 1, -1)

saurabh.hmr009

We use return in programming to tell program execution was successful.Normally 0 stands for successful execution and 1 means there is some error .

vishwajeet.gaur1989

@lkshchoudhary004::

1. Return type in programming is not a compulsion. It depends entirely on your coding logic as to what you want from the functions you created in c. Suppose you are creating a function to add two numbers. Logically the numbers will return int if the two arguments passed to this functions are integer. So you can use int sum(int a,int b) and return a+b.

Now whenever you want the addition of two numbers you can simply assign the addition to new variable using function call like--> c=sum(a,b); Here, sum will return addition of a and b and assign the return value to c.

2.When you come across return types in main function i.e., int main(), usually return 0 is written at the end of your main function to tell the shell that the program has been executed successfully.

Anything apart from 0 is treated as unsuccesfull.

adeshagarwal02

1) return type is used when the function returns some thing it can be any type such as integer char or float depending on prototype of a function for example
int fun() will return integer type value.

2) when int main( ) returns 0 than it mean function runs successfully, and if it returns something else rather than 0 means there is some error in program

lkshchoudhary004

1. Return always store the result of function computation and c is top to bottom procedure language
2. In case of function we need to declare at the top of function define but every time function curly braces hold the
    value
3. This value may be result of some computation so after computation in function resultant store in function by
     return because the function in main program need this result to execute the further program with result.

adeshagarwal02

Is it legal
int main ()
{
int *ptr
*ptr =5;
}
can we do this
I think we can't do this
i need opinion on this

saurabh.hmr009

I think its illegal because pointer always points to the address and that address holds some value in the memory stack , it doesnt store a value .

adeshagarwal02

@saurabh.hmr009
that what i thought
thank you :)

lkshchoudhary004

5 will treated as address not a value

vishwajeet.gaur1989

@adeshagarwal02

@saurabh.hmr009

@lkshchoudhary004

When a variable is declared as pointer type, we need to allocate memory to this pointer pre-assigning it with any values.

*ptr=5 will work only if memory is allocated before assignment.

This program do not posses any syntax error however due to no memory allocated to pointer ptr, this will show runtime error <Segmentation Fault>

adeshagarwal02

char name[ ] ="xyzabcd";
1) printf("%s",name);
or
2) printf("%s",&name);
which one is right 1 or 2
give reason in support of your answer :)

adeshagarwal02

why it is illegal
typedef struct Dis
{
int *feet;
float *inch;
}Dis;

Dis d1;

sankalpnegi2310

@adeshagarwal02

printf("%s",name ) ---->> is valid coz name is the address from where the array is beginning, that is the address of name[0]. where as &name means the address at which the pointer is stored.

sankalpnegi2310

Find the bug in the code:

The code below runs fine when the comment is removed.But it gives segmentation fault when the below code runs.

#include<stdio.h>

int main()
{
        int *y,*p,t=8;
//        p=&t;
        *y=12;

        printf("%p %d %d\n",y,*y,t);

        return 0;
}

sankalpnegi2310

Pointers and linked lists:

(here start is a struct pointer)
1.) Do we need to explicitly put
start->next=NULL;
when we have already done:
free(start)

as when we have freed the given memory location...does it really make any sense to assign a sub instance of the same???

vishwajeet.gaur1989

@sankalpnegi2310 ::

Could you please elaborate more in which context you are writing start->next=NULL when you have you already did free(start).

If free(start) is executed, then there seems to be no point in writing start->next=NULL according to me.

somanshu765

Not able to scan multiple char values

#include<stdio.h>
int main()
{
        char a[3];
        int i;
        for(i=0;i<4;i++)
                scanf("%c",&a[i]);
        return 0;
}

adeshagarwal02

@somanshu765
please leave space in scanf for %c
scanf(" %c",&a[i]);
and please make i<3 in for loop

vishwajeet.gaur1989

@somasnshu765:

@adeshagarwal02:

Leaving space before format specifier is mandatory only when input stream is holding any other data type.

Suppose your input stream holds integer which is 4 bytes long and after this you want to input a character. The program will terminate the statement of scanf when you try to input a character here because scanf was expecting a 4 bytes stream however you gave 1 byte stream.

To avoid this kindly use space. In your case space is not mandatory try to debug the code, it will work without space as you are not using any other data type in input stream apart from char.

saurabh.hmr009

@sankalpnegi2310,
I dont think there is need of using start->next=NULL, when you have already free(start).

somanshu765

@vishwajeet.gaur1989:
@adeshagarwal02:
Referring to my previous quire I found the that our system is behaving in its canonical form as it allows the user to press enter to make the input available to program.
So,Linux is saving the input until the user presses Enter and then pass both the character and subsequent enter to the program.therefore,each time we enter a character scanf scans the char value ,then immediately scans the enter character '\n'(Line feed).

So solution to the problem is:
#include<stdio.h>
int main()
{
        char a[3];
        int i;
        for(i=0;i<3;i++)
        {
                do
                {
                        scanf("%c",&a[i]);
                }while(a[i]=='\n');
        }
        for(i=0;i<3;i++)
                printf("%c",a[i]);
        return 0;
}

sankalpnegi2310

Algorithm for 3-bit compression and decompression :

1) Compression

This will be done in 3 stages each processing a byte individually as 1 byte = 8 bits and we need 3 bits to represent one byte,therefore, total 24 (LCM of 8 and 3 ) bits processing required and thus 3 stages each processing a single byte.
Only after 3 bytes of processing we will be moving to next iteration.
One iteration will be giving us 8 characters encoded/encrypted into 3 bytes

Stage 1:

read a character from file --- find its index in our previously created master array and get the index...which is INTEGER hence 4 bytes
convert into 1 byte using sprintf() (unsigned)
shift it by << 5 to place at msb
similarly place next character read after msb
now read a new character --- place 2 of its bits into the remaining 2 bits..write this byte to file.
we have 1 bit remaining.

Stage 2:

begin a fresh byte ---place remaining bit after having it right shift as << 7
read a new character --- get index ---shift such as to place next to last placed bit.
similarly get new character place it next to last placed bits.
now we have 1 bit left which will be filled by first bit of new character to be read.
read a new char --get index place 1st bit in to the 1 bit left in our byte.
write this byte to the file...we still have 2 bits of this character remaining.

Stage 3:

Get a fresh byte.
place the last 2 bits of previous character in the new byte.
now read 2 more characters from file..and place their indexes one after the other after proper shifting.
we now have our byte complete -- write it to file.

This way we have got our algo for 1 iteration..this shall be put in a loop.. so as to get required no of encypted bytes to be written..

2) Decompression

This shall again be done in 3 stages where each stage will be processing a byte read from our newly created encrypted file. This process is just the reverse of Compression algo as explained above, hence does not require re-explanation.

Note : When implementing both the algorithms we would need a bunch of conditions to check for certain boundary conditions etc and thus put in required statements which have not been covered in here.

However all queries shall be entertained.

adeshagarwal02

I have a query
why i m getting
warning: ‘gets’ is deprecated (declared at /usr/include/stdio.h:638) [-Wdeprecated-declarations]
gets(p);
^
whenever i use gets( )
my program is running perfect but i m getting above warning during compilation

sankalpnegi2310

@adeshagarwal02 The gets() function is no more considered a standard function and has been ruled out, as it causes buffer overflow sometimes..thus the complier tells not to use it or dangerous. The C11 standard ISO/IEC 9899:2011 eliminated gets() as a
standard function,sadly, it will remain in
libraries for many years (meaning 'decades') for reasons of backwards
compatibility.

adeshagarwal02

In the given problem below pow(n,2) is not working
if we do pow(5,2) it will work
#include <stdio.h>
#include <math.h>
int main()
{
        double n,c;
        printf("please enter the number :");
        scanf("%lf",&n);
        c=pow(n,2);
        printf("(%lf)^2=%lf\n",n,c);
        return 0;
}
please try this above code

vishwajeet.gaur1989

@adeshagrwal02

Considering that you are using GCC compiler, beware that gcc and some compilers has pow as built in function for literal constants.

So when you do not use any variable and use constant instead like pow(5,2), the inbuilt pow is called and evaluated( Even if you do not include math.h ).

With a variable input, the compiler must rely on the math library, which you must link against.

To use math.h you need to explicitly link math.h library at the time of compilation as mentioned below.

gcc program.c -lm

adeshagarwal02

Project Description

p { margin-bottom: 0.1in; line-height: 120%; }</</style>In MDC project the
textual data is compressed to reduced file size and to generate
encrypted or compressed file. System calls are used to open and read
the file from which distinct characters are stored in a array which
called master array. From this master array code length of this
distinct characters is found. code length will chose the bit
compression is to start. Now each characters of the textual file is
compared with the master array to generate its index,which is stored
as compressed file and the character of the master array is stored in
an file named as key. Now this key and the compressed file used in
decompression. In decompression the compressed file will provide
index and the key will provide character from the index the
characters are searched from key and put into decompressed file.

Multiple Data Compression and Encryption using Iterative Technique. B97

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