How can i change char to int and send it to massiv in c++

Question

/*
Demak, boshladim
created by A.Behzod
bu dastur sonni shunchaki qabul
qiladi va uni massivga joylaydi

fayl <<bu max ko*paytma>> uchun
maxsus

*/

#include<iostream>
#include<math.h>
using namespace std;

int main()
{

char lg[3];
cout<<endl;

cin >>lg[0]>>lg[1]>>lg[2];//belgilar

for(int i=0; i<10; i++)
{
int son[10]; //massiv
//massivga son kirityapti
son[i] = (int)lg[i] - 48;
cout << son[i];}

main();

return 0;
}

So there is problem that 'i' is changing 0 when code repeat main() function
But i want to send to son[i] integer number from char[lg] one by one

And it is important that
I should press any key durinng three number sended

Answer 1

int son[10];
for(int i=0; i<3; i++)
{
son[i] = lg[i];
cout << son[i]<<endl;
}

This is enough. Behzod
In your code, your are taking 3 inputs but converting 10 charecter (loop 10 times cycling..) and no need subtract a value or type cast (it will type cast implicitly from char to int.. Take initialization of son[10] out off loop...

Answer 2

What is this code? why make main function recursive?

Answer 3

Behzod
Have you got the solution?
I didn't exactly understand the program, maybe it's because I don't understand your language (Uzbek).

Answer 4

Salon
Hello Ipang my code is used for multiplication of two number
Which are inputed as

1)Input->> 2 + enter (first number's digits are 2)

2)continue input->> 2 + enter (second number's digits are 2)

3)continue input->> 1 2 (separate it with probel )

4)continue input ->> 1 3 (separate it with probel)

Output is: 156 ( 12*13=156)

This code works on high digit numbers also

Answer 5

Behzod
24 digit number is very big, I don't know if any integral data type supports that many digits. And being multiplied, it will grow bigger.

Any thoughts on this matter?

Answer 6

#include<iostream>
using namespace std;
int main()

{//beginnig

int x, y;

cout<<"HELLO\nWELLCOME TO \nTHE CODE OF MULTIPLICATION OF VERY LONG\n ";
cout<<"TWO NUMBERS\nXUSH KELIBSIZ¡¡¡\n\n";
cout<<"What is the amount of firs number: ";

cin>>x;
cout<<"\nWhat is the amount of second number: ";
cin>>y;

int a[100][100]={};

cout<<"\nType the number of first one by one using probel \n";

for (int i=1; i<=x; i++)
{ cin>> a[0][i]; }

cout<<"\nType the number of second number one by one as first \n";

for (int i=1; i<=y; i++)
{cin>> a[i][0]; }

cout<<endl<<endl;;

//katakch natijasi
for (int k=1; k<=y; k++)
for (int j=1; j<=x; j++)
a[k][j] = a[0][j]*a[k][0] ;

int b[200][2]={};

cout<<endl<<endl;
for (int k=1; k<=y; k++)
for (int j=1; j<=x; j++)
b[k+j][0]+=a[k][j];

//tuzatish
for (int i=x+y; i>1; i--)
{ b[i-1][0]+=b[i][0]/10;}

//tuzatish

Answer 7

Behzod
DM doesn't work, we discuss here

I'm trying to understand ...
First input defines first number digit
Second input defines second number digit

But why 1 2 and 1 3?
Why not take the number 12 and 13?

Answer 8

Is it ok if we read input as 12 and 13?

I think I begin to understand better ...

Answer 9

Basically you should enter
15
24

1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
And
1 3 5 4 6 7 3 6 2 4 3 5 3 6 4 5 3 4 5 3 3 6 6 3
And you give multiple of them

Answer 10

I might try to help with small number of digits. But 15 ~ 24 digits, I don't know for sure ...

Answer 11

Int first[];

Char dgt[] ;
Cin>>dgt[1]>>dgt[2]>>...>>dgt[n];

For (int i=1; i<=n; i+8+)
{First[i] = dgt[i] - 48; }

This is consept of my problem
But as you see there is another problem

Answer 12

Here it's using `int`
I'm limiting up to 10 digits for each number. `int` can't go more than 10 digits.
Test with input 3 3 1 2 3 4 5 6

https://code.sololearn.com/cp82IKXQUHiP/?ref=app

Answer 13

It is example for understanding how is it used for

Answer 14

But i want input them without probel and as 12 not 1 2
And it will be used more comfortable