Posts by csurfer

One of the problems I see is this

return *output;

It is actually returning output[0] here when you want it to return the address of output array.
It should just be

return output;

Sorry fellows Daniweb is loading too slow on my side so when I was posting I couldn't see your replies... ;)

@OP : If you had put in half of the effort what Vernon has put in the above post you would have solved your problem easily by now !!!

You are not lost after coding this much you are lost from the beginning...

Hell lot of mistakes :

1> [#include <iostream> Typing mistake

2>

int 
int main(void)

What is this form ???

3> cout <<a1[45]<<endl; Segmentation fault illegal memory referenced alloted memory is from a1[0] to a1[44] for declaration int a1[45]; 4>

if (a2[50] % 5 ==0)
cout<<a2[50]<<endl;

Segmentation fault again

5> int max(0); What is this exactly???

6>

if (a1[45] > max)
max = a1[45];

Segmentation fault again.

And USE CODE TAGS and for finding minimum and maximum you can follow this algorithm

int min,max;
//Loop through a1[] first and initialize min=max=a1[0]; Now for every i ahead
if(a1[i]<min) min=a1[i];
if(a1[i]>max) max=a1[i];

//Now loop through a2[] ,for every i
if(a2[i]<min) min=a2[i];
if(a2[i]>max) max=a2[i];

//YOu'll have both min and max values for both arrays combined.

Both of you are right.

Copy constructor is a constructor used only in situations where we come across situations two objects A and B of same class type assigned as

A=B;

I wanted to go with example

X a=X(X const& copyFromMe);

example but it ended up this way.(In making sure I don't give him the thing he seeks in a golden platter ;))

@ramthegreatcv : Ya its return value optimization and about the copy constructor I didn't find any other way to make you understand,so used this bizzare approach.

@tux4life :I was trying to explaining op the purposes of using a copy constructor with that,thats it,I know the way I took is hopeless.

Well how have you written the records in the file from which you are taking the input ?
Extras :
>Try not using .eof() function its not good to use it like this several times discussed already in this forum.
>And within the read looping you can always include a counter right to know how much you have read.

Answers:

1>Assume that there is a class as:

class check{

int a;

public:

check()
{ 
cout<<"Constructor"; 
}

check(int i)
{ 
//copy constructor code 
}

void setvalue()
{ 
//sets value for a 
}
};

//Inside main way 1 normal
check c;
c.setvalue();

//Inside main way 2 copy constructor
check c(5);

Above shown are the two types of initializations for initialising integer a within the class object.
First method is to call the setvalue function as we normally do to set value to that integer.But you can directly initialize the static variables present within the class object as

check c(5);

if you have written code for it properly.

2>Address of bA and aA are same I have answered already why.

3>Value of aInt and bInt cannot be same because you are returning the value of aInt so aInt's value gets assigned to bInt so their values are same but their addresses differ because they are separate variables.

4>I have already explained why the destructor is called at end of main only,its because aA is not destroyed at the end of fun2 its returned and assigned to bA so now after main destructor for this object is called and that is seen by you.

>Why doesn't this happen for primitive data types like int ,char or float ?
I really didn't get your question.But this will give you your answers I suppose. and also this and this.

To be frank I think you don't know the use of copy constructor at all so please read about it first and then ask your doubts if you don't mind.You need to brush up a lot.

@athlon32
Check your answers before posting.

1>

They are not, i ran it on my computer and i get 2 different addresses

The address should be same because the OP is returning the object b address and assigning it to a.
2>

I don't think so, I removed the function fun() and left only a, and it gave me both destructor/constructor messages...i think the des/con aren't being called on B :?

How can you ever get destructor and constructor after removing fun() object b is getting declared within fun().
3>

Because it should be like this:
class A{
int i;
public:
A(){
cout<<"default constructor\n";
}
~A(){
cout<<"default destrutor\n";
}

A(A& obj);
};

and this:
A::A(A& obj)
{
i=obj.i;
cout<<"copy constructor\n";
};

You don't need to just a const suffices look at my post.

@ramthegreatcv
Answers :

>1. Why only 1 destructor is getting called.
Because even though you think you are dealing with two objects you are actually dealing with one,because when you used return b in fun(),what you are actually doing is returning the address of that object and assigning it to object a by the statement a=fun(); so only one object is under observation.Here destructor is called only for object a before returning from main.

>2. Why address of a and b are same.
Look at answer for 1.

>3. If I use copy …

Well I looked at your code and it has a bunch of problems :

Error 1:

If this itself is your final code then you don't need NextDay() and PreviousDay() functions at all.

Error 2:

weekdayT IncrementDay(weekdayT day,int pass){
int p=0;
int counter=0;
p-=pass;
if(pass<0){
if(day>p)
{
return weekdayT(day-p);
}
else
{
counter = day;
for(int i=0;i<p;i++)
{
counter--;
if(counter<1)
{
counter = 7;
}
}
return weekdayT(counter);
}
}
return weekdayT((day%7)+pass);
}

This return should be:

return weekdayT((day+pass)%7);

Error 3:

Your code gives correct result for

cout << "Increment " << IncrementDay(Tuesday,0);

as Increment 0 but for

cout << "Increment " << IncrementDay(Tuesday,-7);

it gives Increment 7 instead of Increment 0 which even though correct is logically wrong.

And ya a general instruction about functions like below if you really use them :

weekdayT NextDay(weekdayT day){
	return weekdayT((day%7)+1);
}

Don't use a function call just for a single statement processing. As you know the processor needs to do a whole lot of stuff before calling a small function like this one too and in cases like this its nothing but waste of processor time.If you can incorporate functions as small as these in macros then it speeds up your program to a good extent.
Though not an error it helps you better your program so …

Hey MosaicFuneral you seem to love spam msgs....thats funny dude !!! :)

Every thing but this makes sense to me :

return weekdayT((day-1)%7);

Assuming that the other user function which you have written is going to send values from 0 to 6 only for a weekday(if you have written that properly;)) then (day-1)%7 makes absolutely no sense and is redundant after what you have checked in your if statement.Only day-1 is enough as %7 there checks a thing you already know(that the value is between 0 and 6).

Well a weird idea but you can even split the number into digits and store in a linked list or an array or something else... flex your brain and you will do great !!!

Many Many Happy Returns Of The Day !!! And Welcome....

>Should I learn how to do C++ in windows and then learn.. Linux shell scripting (did I say that right)?
Well if you ask me thats the easy way but I would suggest you to learn and practice according to standards so install the Ubuntu you have got download a standard version of gcc (for c) and g++ (for C++) and go on coding...

Well I'd like to tell two things :

1> I really didn't get how is your algorithm going to test if there is the thirteenth friday in that month or not.
Your program doesn't have any reference for the year and without that there is no way to check the total number of days from the years start and the computations you have done is something totally out of the way.

2>Here I am not correcting your algorithm but shortening it as you had asked.You can device a far better algorithm so try ;)

//calendar
#include<stdio.h>
#define s scanf
#define p printf
int main ()
{
    int dw,dm,m;
    p("plz enter dw, dm, m\n");
    s("%d%d%d",&dw,&dm,&m);
    //for Alldays taken together

    while (!(dm>=1&&dm<=7)) dm=dm-7;
    
    if(dm==dw) p("in this month there is a 13th friday\n");
    else p("in this month there is no 13th friday\n");

    return 0;
}

You are doing nothing but the above code.And to be frank above code does nothing ;). And ya never use void main,use int main get used to the standards.

Well one of the ways for the first questions as I thought is this.

1> Take an input number n from the user.
2> "x" is an empty variable.
3> You have "y" as a number under concern.

What you need to do is pick the rightmost n bits of y and make a decimal number from it and feed it into x.
So you can do this.
* Assume an integer variable temp=1;
* Run a loop n-1 times and within it do
temp<<1;
temp | 1;
* After this loop you just need to do y & temp.
* The above algorithm will fetch you the right most n bits of y.

So your answer for 2-6 is x = y & temp;

In the same way try to answer 2-7. ;) The above code will make your other bits of "y" as 0. So you take another variable say z = y for your computations so that you ll get the results and also y will remain unchanged.

C runs on some rules,you need to learn them before trying out something,and ya you cannot just translate a code to C and expect it to work.

int maxprimes;
int count = 1;
int value = 1;
int primes[maxprimes];
int composite;
int i = 1;
int j;
printf ("How many primes? ");
scanf ("%d", &maxprimes);

This is wrong. You are trying to take the value of maxprimes from the user and trying to allocate that much memory for the array primes. But as you can see compiler would have already allocated memory for that array and at that time

int primes[maxprimes];

doesn't have any meaning,so the segmentation fault.

What you can do is to take the value of maxprimes and then allocate memory for array,but in C

int maxprimes;
int count = 1;
int value = 1;
int composite;
int i = 1;
int j;
printf ("How many primes? ");
scanf ("%d", &maxprimes);
int primes[maxprimes];

something like this is not allowed.Because in C all data variable's should be declared and initialized before the first operational statement(say an assignment or a check.).Its memory also needs to be allocated priorly if you are allocating it statically.

Now coming to the solution : You need to use dynamic allocation of memory Do this :

//Within main function
int maxprimes;
int count = 1;
int value = 1;
int *primes;
int composite;
int i = 1;
int j;

printf ("How many primes? ");
scanf ("%d", &maxprimes);

//your solution …

> oh thanks man but coudlnt i change direct to buffer?
What directly to buffer ? You mean to say directly to str array right ? I wont say you can't . First you need to write it as you were writing earlier and then after it you need to clear all the memory blocks ahead of it.
That is if your squeezing leaves your str array at index 4 then you need to find the total memory blocks allocated for this array in main function(because thats where you are allocating it memory)and clear all the memory blocks allocated for this block ahead. (Which is the right "FORMAL" procedure...).

Shortcut :
After your str array is over written just move the index 1 block ahead and place '\0' character in it. Placing the '\0' character symbolises the end of string and compiler won't print ahead.

But be warned this method is sure to bring one or the other problem because you won't know where exactly to place the '\0' character.

And ya your program has whole lot of limitations :
1> Your program doesn't take every letter in string2 and delete it in string1.Because its not looping its just matching the indexes. If instead of NAM as second string i give MAN then even with these letters in string1 your code wont delete it because its just matching indexes.

In total :
Try to write a better code for your requisite because …

Well in that case say you have 6 characters in str array,then assuming that you overwrite first three characters in the exact way in which you want even then you are doing nothing with the left over's.
Well change your code as this :

#include <stdio.h>
void Squeeze2(char *str,char *save,char *required)
{
     int i,x;
     int buffer=0;//used to save stuff
// And ya make an habit of using null character rather than simple 0
     for( i=0 , x=0 ; str[i]!='\0' && save[x]!='\0' ; i++, x++ ) {
          if( str[i]!=save[x] ) {
              required[buffer]=str[i];
              buffer++;
              }
          }
}
int main(void)
{
    char name[]="NAME";
    char req[10];
    Squeeze2(name,"NAM",req);
    puts(req);
    getchar();
    return 0;
}

The code is not only weird its senseless. Firstly two pointers buffer and i manipulating the same char array so what it writes in the array in which condition is completely unpredictable it just depends on both the inputs.
And even by going by the function name "squeeze2" if the OP wants to squeeze the first or the second array then why so much redundant work it can even be achieved by inbuilt functions.

Well use getline function. It advances the file get pointer to next line automatically.Or even you can look upon f.seekg() function.

Here is an algorithm for it :

//open a file using fstream objects say file and functions like file.open() in input mode
// Create a string variable which can act as a buffer for holding the values
while(getline( <file descriptor value>, <buffer object>)!=NULL)
{
int key;
string skey,value;
//extract string value of key into skey and convert it into integer using istringstream
//extract string value of key into value
//With the help of a new fstream object write into the file you want it to.
//erase the buffer for providing space for next input using file.erase() function
}

@StuXYZ :

Several things from your question:
First the string part. You have many options (a) you can indeed use
insert but the method requires an iterator to define the position for insertion. (b) the easiest to use simply += e.g

Copy to Clipboard

char p='x';
std::string A("test");
A+=x;
// A is now testx

Make sure that the examples you give are correct because in the above code what you have stated is wrong and it would give rise to a compiler error stating that char variable x not declared.
Its either :

char p='x';
std::string A("test");
A+=p;
// A is now testx

or better and simple do this :

//char p='x';//Not at all required
std::string A("test");
A+='x';
// A is now testx

"After reading all the above posts once" you will know your question itself has your answer.
1>Convert float to string.
2>Concatenate using srtcat or strncat as said by you or snprintf as said by Arkm.

When you know everything did you try to google out first?I think you lost your patience in the last step. !!! ;)

You have a problem with the whole program or just a part of it ? Because its really hard to scan through every part...

Well you get built in options for the things you have asked both in Dev-Cpp and also Code Blocks...Try them.

Well if you can control the output of a random function then it wont remain a RANDOM function right ??? :D
Well you can always do this.First assume a variable "a" which is defined as follows :

1>a=Some random number from 1 to 19 (%20) generated by some random number generating method.
2>Place 0 in this block of array as array[a]=0.
3>Now your are sure that the array has one zero and also in a random place generated by computer itself.
4>Now go on with your for loop as above itself and just skip the loop when i==a.

And ya you have loads of threads here at daniweb to make the random function better.Just have a look at these threads it may help.

I got how you are trying to generate the prime numbers but you have messed it up to a great extent.

>Assume another variable lim which has the maximum index of prime numbers stored in array p.That is if you have got 3 prime numbers p[1],p[2],p[3] then lim is 3 and a variable called flag which signifies divisible if 1 and not divisible if 0.
>Just after your

if (num<5)
for(j=1;j<=((num+1)/2);j++)
printf("%d,",p[j]);
else
{
for(j=1;j<=3;j++)
printf("%d,",p[j]);

step do this :

x=6;
while(x<num)
{
           flag=0;//Signifies not divisible
           for(i=1;i<=lim;i++)
          {
                      if(x%p[i]==0)
                     {
                               flag=1; //shows its divisible
                               break;
                     }
         }
         if(flag==0)
         {
                      p[++lim]=x;
         }
         x++;
}
//Print the array p[] from 1 to lim here and you are done.

Its just this much even if I follow your algorithm.You are just confused.Take time to view your code once and the optimized code given above.And ya put your codes within code tags from next time.

I suppose you are using some compiler like TC because if you try to do something like array[-1] in gcc then it gives a segmentation fault.
Actually the concept is like this.When you write a statement like

int array[10];

10 consecutive integer blocks are alloted in memory with name array which can be accessed as array. When you write array[-1] then it tries to access an integer block before the memory allocated for array.Hence is illegal memory access.Hence not a valid statement.

And

array[1] = *(array+1);

because as you know

array=&array[0]

,you can access the 2nd block of memory allocated for array or the [1]th block as array[1] or *(array+1) because

*(array+1)=*(&array[0]+1)

which means get the array starting address increment it by 1 so it will point to second block of memory then get its value by * operator.

Think this can help you a bit.

Use getline(<fileptr>,<string/buffer to read the line into>,<delimiter>);