Metacafe.com

If u r much interested in watching videos from across the world and wanna save them in ur PC?

here is the useful software that will make ur complete work.



have a look at this site once. Metacafe

It automatically downloads many videos from the internet when ur internet connection is idle.

There are option to specify ur interested topics and options to control the type of videos that has to be downloaded also.

U need not worry, if u r at home u can control many videos not being downloaded.



Its really cool one i got.

Slideshare.com

A good website for sharing powerpoint presentations. Slideshare

While browsing the net, i got this site. I felt very interesting....



Try this once. Hope u love it. It needs flash payer installed in the browser.

Complete text of the presentation is available in the bottom of the page. It is awesome. initially, We can view the text before watching entire presentation.

http://www.slideshare.net/jignesh/ddos-attacks

Question for the week -2

2 power 4 = 16 and the sum of the digits is 1 + 6 = 7

2 power 15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26



What is the sum of the digits of the number 2 power 1000

Questions for the week - 1

//correct all errors and make the program do insertion sort



// There are some logical errors. Note that, there is no compilation error.



// program compiled in gcc linux fedora core 5







int X[10], 



   Y[10], 



   NumInputs, 



   NumY = 0; 



             



void GetArgs(int AC, char **AV)







{  int I;











  NumInputs = AC - 1;



  for (I = 0; I  NumInputs; I++)



     X[I] = atoi(AV[I+1]);



}







void ScootOver(int JJ)







{  int K;







  for (K = NumY-1; K  JJ; K++)



     Y[K] = Y[K-1];







}







void Insert(int NewY)







{  int J;







  if (NumY = 0)  {



                  



     Y[0] = NewY;



     return;



  }







  for (J = 0; J  NumY; J++)  {



     if (NewY  Y[J])  {







       



        ScootOver(J);



        Y[J] = NewY;



        return;



     }



  }



}







void ProcessData()







{ 







  for (NumY = 0; NumY  NumInputs; NumY++)



    



     Insert(X[NumY]);







}







void PrintResults()







{  int I;







  for (I = 0; I  NumInputs; I++)



     printf("%d\n",Y[I]);



}







int main(int Argc, char ** Argv)







{ 







  GetArgs(Argc,Argv);







  ProcessData();



  PrintResults();



}

Quine

Write a C program that executes and prints itself in the output.

(Do not use any file concepts such as reading from source file, the source will not be provided.)

The answer is
A program that produces its complete source code as its only output is called a quine. This is a technique some times used in viruses.( dont ask me to write a virus or to help in writing a virus ).

char*f="char*f=%c%s%c;main()

{printf(f,34,f,34,10);}";
main(){printf(f,34,f,34,10);}

Let us understand the above program.
Before that, have a look at this code
char t[10]="%s";
char name[10]="hello";
printf(t,name);

it prints
hello

generally printf takes parameters. treats the first string is to be printed and if there is some %s in the first string, it searches for the next parameter(ignores it, if there is no corrosponding parameter in the printf call). prints it with out any doubt.

Now, let us understand this code

The program is
char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}";
main(){printf(f,34,f,34,10);}

initially, we r declaring a character pointer f.
we are storing the text that is present in the program into the pointer p but, with some technique to print itself twice.

char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}";
main(){printf(f,34,f,34,10);}

we are storing char *f=%c%s%c;main(){printf(f,34,f,34,10);}
in to the character pointer f
in the main we are calling a printf.
f is passed as a first variable to printf

char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}";
main(){printf(f,34,f,34,10);}

The printf will look like

printf("char*f=%c%s%c;main(){printf(f,34,f,34,10);}" ,34,f,34,10);
observer the double quotes text in printf, it has %c%s%c

printf("char*f= %c%s%c;main(){printf(f,34,f,34,10);}",34,f,34,10);

there i am printing 34, f, 34. (34 is ascii double quotes ("))
the entire printf will look like
printf("char*f=" char*f=%c%s%c;main(){printf(f,34,f,34,10);}"",10);

we can omit 10. it is not necessary.

if it is not clear, write the program on a paper and try the steps described above.

if u r comfortable, have a look at ............(not compulsary, leave this complex example if u dont understand)
A complex quine example

/* A simple quine (self-printing program), in standard C. */

/* Note: in designing this quine, we have tried to make the code clear and readable, not concise and obscure as many quines are, so that the general principle can be made clear at the expense of length. In a nutshell: use the same data structure (called "progdata" below) to output the program code (which it represents) and its own textual representation. */

#include

void quote(const char *s)
/* This function takes a character string s and prints the
* textual representation of s as it might appear formatted in C
* code. */
{
int i;

printf(" \"");
for (i = 0; s[i]; i++) {
/* Certain characters are quoted. */
if (s[i] == '\\')
printf("\\\\");
else if (s[i] == '"')
printf("\\\"");
else if (s[i] == '\n')
printf("\\n");
/* Others are just printed as such. */
else
printf("%c", s[i]);
/* Also insert occasional line breaks. */
if (i % 48 == 47)
printf("\"\n \"");
}
printf("\"");
}

/* What follows is a string representation of the program code, from
* beginning to end (formated as per quote() function, above), except
* that the string _itself_ is coded as two consecutive '@'
* characters. */
const char progdata[] =
"/* A simple quine (self-printing program), in st"
"andard C. */\n\n/* Note: in designing this quine, "
"we have tried to make the code clear\n * and read"
"able, not concise and obscure as many quines are"
", so that\n * the general principle can be made c"
"lear at the expense of length.\n * In a nutshell:"
" use the same data structure (called \"progdata\"\n"
" * below) to output the program code (which it r"
"epresents) and its own\n * textual representation"
". */\n\n#include \n\nvoid quote(const char "
"*s)\n /* This function takes a character stri"
"ng s and prints the\n * textual representati"
"on of s as it might appear formatted in C\n "
"* code. */\n{\n int i;\n\n printf(\" \\\"\");\n "
" for (i = 0; s[i]; i++) {\n /* Certain c"
"haracters are quoted. */\n if (s[i] == '\\\\"
"')\n printf(\"\\\\\\\\\");\n else if ("
"s[i] == '\"')\n printf(\"\\\\\\\"\");\n "
" else if (s[i] == '\\n')\n printf(\"\\\\n\""
");\n /* Others are just printed as such. *"
"/\n else\n printf(\"%c\", s[i]);\n "
" /* Also insert occasional line breaks. */"
"\n if (i % 48 == 47)\n printf(\"\\"
"\"\\n \\\"\");\n }\n printf(\"\\\"\");\n}\n\n/* What "
"follows is a string representation of the progra"
"m code, from\n * beginning to end (formated as pe"
"r quote() function, above), except\n * that the s"
"tring _itself_ is coded as two consecutive '@'\n "
"* characters. */\nconst char progdata[] =\n@@;\n\nin"
"t main(void)\n /* The program itself... */\n{\n"
" int i;\n\n /* Print the program code, chara"
"cter by character. */\n for (i = 0; progdata[i"
"]; i++) {\n if (progdata[i] == '@' && prog"
"data[i + 1] == '@')\n /* We encounter "
"two '@' signs, so we must print the quoted\n "
" * form of the program code. */\n {"
"\n quote(progdata); /* Quote all. *"
"/\n i++; /* Skip second"
" '@'. */\n } else\n printf(\"%c\","
" progdata[i]); /* Print character. */\n }\n "
" return 0;\n}\n";

int main(void)
/* The program itself... */
{
int i;

/* Print the program code, character by character. */
for (i = 0; progdata[i]; i++) {
if (progdata[i] == '@' && progdata[i + 1] == '@')
/* We encounter two '@' signs, so we must print the quoted
* form of the program code. */
{
quote(progdata); /* Quote all. */
i++; /* Skip second '@'. */
} else
printf("%c", progdata[i]); /* Print character. */
}
return 0;
}


I hope the things are clear about quine. if any one can explain in better way, mail me i will forward it to all.
Thank you.

C - Review II

1. struct fprintf{

int i;
};
void main(){
int fprintf;
struct fprintf cprintf;
cprintf.i=55;
printf("%d",cprintf.i);
}
Output : 55

2. main(){

int a = 88;
printf("%d",a<<3);
printf("%d",a<<1);
printf("%d",a<<1);
}
what is the value of a after the execution of 3 printf statements

Answer : 88. We are not modifying a value

C - Review II 3

3. Explain which one is better and why? or are they same? There is no compilation or runtime error

i) int a[1000][1000];
for(int i =0;i < 1000;i++)
for(int j=0;j < 1000;j++)
printf(“%d”,a[i][j]);

ii) int a[1000][1000];
for(int j = 0;j < 1000;j++)
for(int i = 0;i < 1000;i++)
printf(“%d”,a[i][j]);

Answer : Code snippet (i) recommonded when programming language uses row-major order

Code snippet (ii) recommonded when programming language uses column-major order

C - Review II 4 ,5

4. main() {

int reset(int );
int i=5;
reset(i);
}
int reset(int n){
if(n==0)
return 0;
else
printf(“%d”, n);
reset(n--);
}
Output: infinite recursion. It continuously prints 5 and after some time stack overflow occurs.

5.

main(){

register int i;

int *ptr;

ptr = &i;

i=10;

printf(“%d”,*ptr);

}

Output: Compilation error. We cannot refer the address of a register variable

C - Review II 6,7

6. What's the "condition" so that the following code snippet prints both HelloWorld !

if "condition"
printf ("Hello");
else
printf("World");

Answer: “condition” can replaced by !printf(“Hello”);
or printf(“Hello”)&&0
or printf(“Hello”), 0
Any expression with commas will take the final value as the expression value.

7. main(){

int i=5;

printf(“%d”,i++ * i++);

i = 5;

printf(“%d”,++i * i++);

i = 5;

printf(“%d”,i++ * ++i);

i=5;

printf(“%d”,++i * ++i);

}

Ouput: 25 36 36 49

When a variable is modified more than once in an expression it takes same value(infact final value) for all the uses or that variable.

C - Review II 8 - 11

8. main(){

int i;

i=1,2,3,4;

printf(“%d”,i);

}

output: 1
i=1 is one expression. After that 2, 3, 4 are taken as three statemets. C will ignore those three.

9. Is this code correct? If not, suggest possible modifications.

main(){

int i,j;

i=10; j=20;

i > j ? i : j = 0;

printf(“i=%d, j=%d”,i,j);

}

Output: Lvalue required

Modify the expression as *(i > j ? i : j) = 0;

The larger one will be assigned to zero.
10. main(){

int a[10],i;

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

scanf(“%d”&a[i]);

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

printf(“%d”,i[a]);

}

input : 1 2 3 4 5 6 7 8 9 1

output : 1 2 3 4 5 6 7 8 9 1

a[i] and i[a] are similar


11. main(){

char *p;

gets(p);

puts(p);

}

Output: segmentation fault.

We are using the pointer without allocating memory