Sure, 1btc per hour and we will all teach you

I don't mean to be rude,,, but why?

It is a good question for me because a wanna be a programmer

sure, but why should others suffer because of you?

Is that a problem if i will be a programmer for others or do you wanna say why others teach you without any payment?

i don't know what are your sources

In other way if you dont want to teach me please send me links which can be sources of coding videos in private inbox chat

Hello I wanted to ask if it’s possible to post a c# problem that I have been battling with

Hello I wanted to ask if it’s possible to post a c# problem that I have been battling with

no, write to C# group, it is in the rules...

did you try Csharp

i dont get you

private void btn_Click(object sender, EventArgs e) { this.Hide(); salesitems op = new salesitems(); op.Show(); int i = 0; Button btn = sender as Button; string btntext = btn.Text; kk.getunpaid(int.Parse(btn.Text)); op.dataGridView2.DataSource = kk.getunpaid(int.Parse(btn.Text)); foreach (var dy in kk.getunpaid(int.Parse(btn.Text))) { op.dataGridView2.Rows[i].Cells["quantity"].Value = dy.quantity.ToString(); op.dataGridView2.Rows[i].Cells["Tprice"].Value = dy.tprice.ToString(); i++; op.libtabsno.Text = dy.tabel_sno_.ToString(); op.libuser.Text = dy.user.ToString(); op.libtno.Text = dy.tabel_no_.ToString(); op.libt2.Text = dy.total.ToString() And this to open unpaidchecks But when I try to add more items a new row is not added to the database but instead it updates the last row.

you better make a screenshot, code like this is unreadable

at least apply monospace to it

Hi everyone, I'm newbie to C language. I'm trying to learn and implement a recursive binary search into my program, to search words into a dictionary. I have a struct struct dizionarioStruct { struct dizionarioStruct *sinistra, *destra; char parola[10], significato[20], sinonimo[20]; } *root = NULL; typedef struct dizionarioStruct dizionario; This is my function, which doesn't work, Im not sure if it is correct nor I call it correctly int binarycercaParola(dizionario * dictio, int l, int r, char parola[]) { if (r >= l) { int mid = l + (r - l) / 2; if (dictio[mid].parola == parola) return mid; if (dictio[mid].parola > parola) return binarycercaParola(arr, l, mid - 1, parola); return binarycercaParola(arr, mid + 1, r, parola); } return -1; } Could you please help me? That's how I call it int risultato = binarycercaParola(ptr->parola, ptr->sinistra, ptr->destra, w); Where w is the word I digit. Thanks in advance

two main issues i found here (along other few things): First, you dont check for NULL nodes Second, you dont use strcmp to compare strings (char vectors) you are not moving through a tree. You made a function to move through a vector of nodes, using l and r as indexes like it were a heap (tree behaving list). That's what the sinistra i destra are for: to move through the tree recursively, considering it's a BST and that elements are inserted following alphabetic order; reason why you are trying to compare chars as ints. Third, your function doesnt return a node nor a string. It's a int and its name as variable isnt what it's supposed to mean Fourth, you are passing dizionario pointers parameters to int arguments

int find(Key key, Node root){ while(root != null && root.key != key){ if(key < root.key) root = root.left; else(key > root.key) root = root.right; } if(root == null) return null; return root.data; } this is a more basic example which makes correct use of the pointers; adapt it so it compares the strings correct position, keeping track of latest error so in case you reach the last attempt of match you get to return -1. You'll prob only need 3 arguments, those being currentNode, currentindex for comparison (so you keep track of which char in the key and in the current node held word are to be compared) and finally the key (or searchedWord). Have fun and I hope I'll see you in compilers course later. https://media.geeksforgeeks.org/wp-content/uploads/BST.png

int find(Key key, Node root){ while(root != null && root.key != key){ if(key < root.key) root = root.left; else(key > root.key) root = root.right; } if(root == null) return null; return root.data; } this is a more basic example which makes correct use of the pointers; adapt it so it compares the strings correct position, keeping track of latest error so in case you reach the last attempt of match you get to return -1. You'll prob only need 3 arguments, those being currentNode, currentindex for comparison (so you keep track of which char in the key and in the current node held word are to be compared) and finally the key (or searchedWord). Have fun and I hope I'll see you in compilers course later. https://media.geeksforgeeks.org/wp-content/uploads/BST.png

I created a file containing multiple lines using \n. But when reading from it, the compiler showed the content in one line not separate lines.

I couldn’t find any c# group

if you use iterations it wont be a recursive function anymore

Print the employee name and joined month in work Steven january

What's the last time the group had a useful discussion?

.

What's the last time the group had a useful discussion?

maybe later, now we are busy with our home assignments

Can anyone here help me i c program for my exam plz

Can anyone here help me i c program for my exam plz

https://twitter.com/mcypark/status/1352022455846506497?s=19

why do i see haskell

Who can help me I'll pay 🙏🙏🙏🙏 🔼🔼🔼🔼🔼🔼🔼🔼🔼 Plz help

pastebin.com/qVRNQdzm

# include <iostream> using namespace std; int main(){     cout<<"Enter the length of array"<<endl;     int n, counter, length, cd=0;     cin>>n;     int arr[n];     cout<<"Enter the elements of array"<<endl;     for(int i = 0; i<n ; i++){         cin>>arr[i];     }     for(int i=0; i<n; i++){         cd = arr[i+1]-arr[i];         for(int j =i; j<n; j++){             if (arr[j+1]-arr[j]==cd){                 counter++;             }             else{                 break;             }         }         if(counter>length){             length=counter;         }         counter =0;     }     cout<<length<<endl;     return 0; }