i have to save,close and open for that?
No
Евгений
Gilded
Евгений
It should indent your code
Anonymous
#include <stdio.h>
int dost();
int main() {
dost();
return 0;
}
int dost() {
int a, b;
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a number: ");
scanf("%d", &b);
return printf("%d + %d = %d", a, b, (a + b));
}
Can you explain lit diffrence between my code to this how it is working???
admin
// Decimal to Hexa Conversion
#include <bits/stdc++.h>
using namespace std;
int dec2hex(int dec)
{
// dec=5386
int a, b, ans;
int i = 0;
string str[10];
char ch;
while (a > 0)
{
a = dec / 16; // 336,21,
b = dec % 16; // 10,0
str[i] = b;
dec = dec / 16; // 336
i++;
}
int x = str[10].size();
for (int j = x - 1; j >= 0; j--)
{
cout<<str[j];
}
}
int main()
{
int dec;
cin >> dec;
cout << dec2hex(dec) << endl;
}
admin
code is not working... plz help anyone.
Евгений
Can you explain lit diffrence between my code to this how it is working???
I don't know how it works))
Ludovic 'Archivist'
Why is that?
Because if you want to use the native APIs or make drivers or use DirectX comfortably it is the choice on Windows
Евгений
You did copy to some where?
Хаха, of course not. I even snowed you two more ways of doing it.
admin
// Decimal to Hexa Conversion
#include <bits/stdc++.h>
using namespace std;
int dec2hex(int dec)
{
// dec=5386
int a, b, ans;
int i = 0;
string str[10];
char ch;
while (a > 0)
{
a = dec / 16; // 336,21,
b = dec % 16; // 10,0
str[i] = b;
dec = dec / 16; // 336
i++;
}
int x = str[10].size();
//print reverse the number
for (int j = x - 1; j >= 0; j--)
{
cout<<str[j];
}
}
int main()
{
int dec;
cin >> dec;
dec2hex(dec) ;
}
admin
code is not working... plz help anyone.
Anonymous
Why did declare out of function?
Anonymous
Just explore what I've sent you and you'll get the idea
I want know some explanation
Danya🔥
Because if you want to use the native APIs or make drivers or use DirectX comfortably it is the choice on Windows
Is there any problems with Clang?
It can even use Microsoft STL as standard lib
Danya🔥
Is there any problems with Clang?
It can even use Microsoft STL as standard lib
For example, Chrome is built with clang on Windows
Emmanuel
include<stdio.h>
int dost(int a, int b, int c);
int main()
{
dost();
return 0;
}
void dost(int a, int b, int c) {
printf ("Enter first number\n");
scanf("%d", &a);
printf ("Enter second number\n");
scanf("%d", &b);
c = a + b;
printf("The sum is : %d", c);
}
The problem is not with your return type. The error is, you created a function prototype "int dost" which takes in three arguments. Now, you're calling the same function "dost()" in your main function without supplying it with any parameter. The compiler will flag an error because "dost()" doesn't have enough arguments as prescribed in the prototype.
Ludovic 'Archivist'
Is there any problems with Clang?
It can even use Microsoft STL as standard lib
Integration, for the rest clang is rather better
Anonymous
#include <stdio.h>
int dost();
int main() {
dost();
return 0;
}
int dost() {
int a, b;
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a number: ");
scanf("%d", &b);
return printf("%d + %d = %d", a, b, (a + b));
}
So finnally its solved by self with Lil help understand by with this code
Anonymous
#include <stdio.h>
int dost();
int main()
{
dost();
return 0;
}
int dost() {
int a, b,c;
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a number: ");
scanf("%d", &b);
c = a+b;
printf ("The sum is : %d", c);
return c;
Anonymous
This is the right code
Marcio
// Decimal to Hexa Conversion
#include <bits/stdc++.h>
using namespace std;
int dec2hex(int dec)
{
// dec=5386
int a, b, ans;
int i = 0;
string str[10];
char ch;
while (a > 0)
{
a = dec / 16; // 336,21,
b = dec % 16; // 10,0
str[i] = b;
dec = dec / 16; // 336
i++;
}
int x = str[10].size();
//print reverse the number
for (int j = x - 1; j >= 0; j--)
{
cout<<str[j];
}
}
int main()
{
int dec;
cin >> dec;
dec2hex(dec) ;
}
You have to treat the hexa digits greater than 9 (transform 10 to A, 11 to B, etc.)
𝕷𝖔𝖗𝖊𝖓𝖟𝖔
Hi guys, so I was given the task to write a software that takes nBytes random bytes from /dev/random device (let it be fd2, after a read only open invocation), and writes it on another file opened in write, let it be fd1. This is quite easy, yet there's a complication: fd1 must be filled with shorts (so 2 bytes) . My solution was to read 2 bytes per time (untill I get to nBytes), thus filling a buffer like the one following:
char buff[2];
and then writing those 2 bytes on the new file. The question is about what happens while writing: so, I have those 2 bytes written in memory, but will they be considered as a short? Or just two random bytes separated? In class we were only told that a short is 2 bytes, nothing further about memory representation, so I want to make sure that the two bytes I write really are a short, and the next two bytes I write are considered as another individual short. Thanks!
Following, the part of the code that does the read from fd2 -> write to fd1 job. I'll quickly resume who's who:
1) fd1 is the destination file (the one I want to write on);
2) fd2 is the source file (the one I opened in reading to get the bytes from);
3) nBytes is the number of bytes (given by the user, I made sure at the beginning of the code that it always is a multiple of 16) the users wants to be written on the file. Notice: not the number of shorts, but the number of the bytes. For istance, if the user wants 2^15 bytes, then, given the fact that a short is 2 bytes, I shall write 2^14 shorts;
4) counter is, as the name suggests, a counter I use in order to understand how many bytes were already copied in fd1.
// Rest of the code was not posted because it is not relevant:
char buff[2];
unsigned long int counter= 0;
do{
if(read(fd2, buff, counter) != 2){
printf("Error in reading!\n");
exit(1);
}
if(write(fd1,buff, 2) != 2){
printf("Error in writing!\n");
exit(1);
}
counter += 2;
} while (nBytes - counter >= 0);
Anonymous
Hi guys which youtube channel is good for c language
Anonymous
A channel?
Nomid Íkorni-Sciurus
A channel?
sorry, it had to be a joke, I couldn't resist.
youtube and learning do not go well together in this field
Nomid Íkorni-Sciurus
what I meant is that you should look for better resources on the web
redfox
can anyone explain templates to me
Nomid Íkorni-Sciurus
can anyone explain templates to me
Imagine you have a Box
a Box can contain anything
so how can you say two Boxes are equal if they're content is unknown?
one thing the templates are used for, is specifying this kind of relationship so that Box<A> and Box<B> will be different but Box<A> and Box<A> will be equal
Nomid Íkorni-Sciurus
this is just one of the applications though.
they're a vast concept I advice you to dig deeper
Faramarz
Hi can you help me to figure out how to avoid memory leak when I'm using malloc?
Thank you
$ameer
Can someone please guide me pops Nasasity
Nomid Íkorni-Sciurus
Hi can you help me to figure out how to avoid memory leak when I'm using malloc?
Thank you
heh you'll need to track your allocations I'm afraid
there are tools that will help you but the best advice is always...
check that your free()s match your malloc()s
Anonymous
Precisely it is not
Anshul
/*
* Below is the interface for Iterator, which is already defined for you.
* DO NOT modify the interface for Iterator.
*
* class Iterator {
* struct Data;
* Data* data;
* public:
* Iterator(const vector<int>& nums);
* Iterator(const Iterator& iter);
*
* // Returns the next element in the iteration.
* int next();
*
* // Returns true if the iteration has more elements.
* bool hasNext() const;
* };
*/
class PeekingIterator : public Iterator {
public:
const int* it;
int s;
const int *a;
PeekingIterator(const vector<int>& nums) : Iterator(nums) {
// Initialize any member here.
// DO NOT save a copy of nums and manipulate it directly.
// You should only use the Iterator interface methods.
it=&nums[0];
s=nums.size();
a=&nums[0];
}
// Returns the next element in the iteration without advancing the iterator.
int peek() {
return *it;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() {
int const temp=*it;
it=it+1;
return temp;
}
bool hasNext() const {
if((it-a)>=s)
return false;
return true;
}
};
can anyone help why am i getting a run time error? i have checked it a few times but i can't find any error
Que: https://leetcode.com/problems/peeking-iterator/
Anshul
the logic that i am applying is: i am keeping an iterator to start of vector,
whenever peek is called i just return what's pointed by iterator.
for next function, i store what is pointed by iterator, increment the iterator to the next step and then return the stored value.
and for hasnext function, i check if it-(starting address of vector) is greater than or equal to vector size, if yes there are no more elements avlbl return false else return true
CALVIN
How to filter out a record of a file maybe using one variable....... But let me share my code first where I attempted this....
Natanim
im trying to write the output of a code but its hard unless i show it by an image
i have a question about it
Natanim
its when outputting a number pattern in a diamond form, but the pattern should look something like this»
> 3
232
12321
232
3
Natanim
its when outputting a number pattern in a diamond form, but the pattern should look something like this»
> 3
232
12321
232
3
/*this code outputs from the outside in but im having trouble shifting it so it could output the above one, i need some help here*/
#include <iostream>
using namespace std;
int main() {
int i, j, k, l, rows;
cout << "Enter Diamond Number Pattern Row = "; cin >> rows;
cout << "Diamond Number Pattern\n";
for(i = 1; i <= rows; i++) {
for(j = 1; j <= rows - i; j++)
{ cout << " "; }
for(k = i; k >= 1; k--)
{ cout << k; }
for(l = 2; l <= i; l++)
{ cout << l; }
cout << "\n"; }
for(i = rows - 1; i > 0; i--)
{ for(j = 1; j <= rows - i; j++) { cout << " "; }
for(k = i; k >= 1; k--)
{ cout << k; }
for(l = 2; l <= i; l++)
{ cout << l; }
cout << "\n"; }
return 0;
}
Pita
#include <iostream>
using namespace std;
main(){
int data, jumlah, cacah;
jumlah = 0;
data = 0;
cacah = 0;
while (data != -1)
{
cout << "Masukkan data angka : ";
cin >> data; jumlah += data; cacah++;
}
cout << " Jumlah data adalah : " << jumlah << endl;
cout << "Rata-rata :" << jumlah/cacah;
}
三体183号
You are too voluptuous
Pita
三体183号
I used to be satisfied with learning for two hours, but now it's painful not to learn for two hours
三体183号
You can see their changes every time
Anonymous
/*
* Below is the interface for Iterator, which is already defined for you.
* DO NOT modify the interface for Iterator.
*
* class Iterator {
* struct Data;
* Data* data;
* public:
* Iterator(const vector<int>& nums);
* Iterator(const Iterator& iter);
*
* // Returns the next element in the iteration.
* int next();
*
* // Returns true if the iteration has more elements.
* bool hasNext() const;
* };
*/
class PeekingIterator : public Iterator {
public:
const int* it;
int s;
const int *a;
PeekingIterator(const vector<int>& nums) : Iterator(nums) {
// Initialize any member here.
// DO NOT save a copy of nums and manipulate it directly.
// You should only use the Iterator interface methods.
it=&nums[0];
s=nums.size();
a=&nums[0];
}
// Returns the next element in the iteration without advancing the iterator.
int peek() {
return *it;
}
// hasNext() and next() should behave the same as in the Iterator interface.
// Override them if needed.
int next() {
int const temp=*it;
it=it+1;
return temp;
}
bool hasNext() const {
if((it-a)>=s)
return false;
return true;
}
};
can anyone help why am i getting a run time error? i have checked it a few times but i can't find any error
Que: https://leetcode.com/problems/peeking-iterator/
The code looks fine and would work if the vector is not updated after being passed to the PeekingIterator constructor. A problem may arise because the question explicitly asks you not to save nums and work only with the interface. So suppose say something is added to the vector after being passed to the iterator then your pointers may be invalidated which may be causing the problem.
You should write peek using Iterator::next and Iterator::hasNext.
Anshul
The code looks fine and would work if the vector is not updated after being passed to the PeekingIterator constructor. A problem may arise because the question explicitly asks you not to save nums and work only with the interface. So suppose say something is added to the vector after being passed to the iterator then your pointers may be invalidated which may be causing the problem.
You should write peek using Iterator::next and Iterator::hasNext.
But where am I saving nums, I'm accessing it using a pointer to a const int
Anonymous
But where am I saving nums, I'm accessing it using a pointer to a const int
Yes but if elements are added to the vector after calling your constructor, these pointers may become invalid
Anonymous
The calling code could be like
vec<int> tmp;
PeekingIterator it(tmp);
tmp.push_back(.......);
Anshul
The calling code could be like
vec<int> tmp;
PeekingIterator it(tmp);
tmp.push_back(.......);
Oh yeah I understand it now, it's going to give wrong answer even if it executed
but atleast the code should produce some output rather than giving a run time error
Anonymous
Oh yeah I understand it now, it's going to give wrong answer even if it executed
but atleast the code should produce some output rather than giving a run time error
It is Undefined Behavior to access invalidated pointers. In all likelihood you may be getting a segmentation fault which is what is being probably shown to you as a RTE by whatever onlineplatform you are coding on.
Anshul
It is Undefined Behavior to access invalidated pointers. In all likelihood you may be getting a segmentation fault which is what is being probably shown to you as a RTE by whatever onlineplatform you are coding on.
Can you please tell me how it is a invalidate pointer?
Like even if the pointer points to a garbage and I try to print it, it'll print that garbage
P.s. I never came across the term invalidate pointer😅
Anonymous
Can you please tell me how it is a invalidate pointer?
Like even if the pointer points to a garbage and I try to print it, it'll print that garbage
P.s. I never came across the term invalidate pointer😅
No. It is Undefined Behavior according to the standard. The C++ standard doesn't and can't define what happens when you access invalid memory. An OS/HW may just page fault. Another OS might crash. Yet another OS might just print garbage values. This is precisely what Undefined Behavior means
Anshul
Thanks I understood this
And while going through solution to this problem I came across this way to write peek
Return (*this).next
Anshul
int peek()
{
return Iterator(*this).next();
}
How does this work?
Anonymous
int peek()
{
return Iterator(*this).next();
}
How does this work?
The way it is supposed to work. Though it is a tad inefficient but. You can implement it a bit more directly.
Iterator(*this) creates a new iterator that shares state with your current iterator pointed to by this. You call next on it which returns the element to be peeked at. The temporary is destroyed after that. The original object pointed at by "this pointer" is not changed.
Anshul
I'm trying to understand this
Anonymous
I'm trying to understand this
It is like this
vec = {1,2,3};
Iterator a(vec);
Iterator b(a);
b.next(); //will print 1
a.next(); //will still print 1
Anshul
The way it is supposed to work. Though it is a tad inefficient but. You can implement it a bit more directly.
Iterator(*this) creates a new iterator that shares state with your current iterator pointed to by this. You call next on it which returns the element to be peeked at. The temporary is destroyed after that. The original object pointed at by "this pointer" is not changed.
There is an object made of peekingIterator, which calls peek
Now this will make a object of iterator class and call it's next function
And this is pointing to peekingIterator object??
Anshul
It is like this
vec = {1,2,3};
Iterator a(vec);
Iterator b(a);
b.next(); //will print 1
a.next(); //will still print 1
b and a both are pointing to same memory or a deep copy is created here 😐
Anonymous
b and a both are pointing to same memory or a deep copy is created here 😐
That depends on how Iterator class is implemented. In all likelihood they will just have copies of the state they are in i.e. the element they are pointing at.
So calling next on on one iterator does not affect the state on the other iterator.
And you shouldn't be worried about it either. The question says that Iterator has a copy constructor and so you can assume it is implemented correctly and just use it
Anshul
I am able to understand it now
Thank you so much
Anshul
The way it is supposed to work. Though it is a tad inefficient but. You can implement it a bit more directly.
Iterator(*this) creates a new iterator that shares state with your current iterator pointed to by this. You call next on it which returns the element to be peeked at. The temporary is destroyed after that. The original object pointed at by "this pointer" is not changed.
Btw it's inefficient because we're calling copy constructor?
And what can be the efficient way then
klimi
(tho if you are working with int you don't have to worry about anything as it is such small size)