draw X letter shape using asterisk(*) - c++

Question

i want to write a program to draw the shape of X letter using asterisk(*)
#include "stdafx.h"
#include <iostream>
using namespace std;
int main(int argc, char* argv[]){
int i, j;
for(i = 1; i <= 9; i++){
for(j = 1; j <= 12; j++){
if(i == j){
cout << "***";
}else{
cout << " ";
}
}
cout<< endl;
}
return 0;
}
i am very new to programming
i only made (\) how can I make the whole X
***------***
-***----***-
--***--***--
---******---
--***--***--
-***----***-
***------***
that's what i did uptill now
include "stdafx.h"
#include <iostream>
using namespace std;
int main(int argc, char* argv[])
{
int i, d, a=1,b=12,c;
for(i = 1; i <= 6; i++)
{
for (d=1; d<i;d++) {cout <<" ";}
cout<<"***";
for (c=a+1; c<b;c++) {cout <<" ";}
{cout<<"***";}
for(b=12-i;b<i;b++)
{cout<<"***";}
cout<<endl;
a++;
}
return 0;
}
i divided the top of the (\//) to three parts
[space][][space][]

Answer 1

I have written the following function/method in java. You can convert it to c++;
public static void printX(int x) {
char[] chars = new char[x];
for (int i = 0; i < x; i++) {
chars[i] = '*';
chars[x - 1 - i] = '*';
for (int j = 0; j < x; j++) {
if (j == i || j == (x - 1 - i)) {
continue;
}
chars[j] = ' ';
}
System.out.println(new String(chars));
}
}
If you call the above function/method as printX(5); The output will by 5x5 sized and containing X character.
* *
* *
*
* *
* *

Answer 2

**** ****
*** ***
** **
* *
** **
*** ***
**** ****
Firstly, pardon my very uneven X. For you as a beginner I would give out an algo for you to ponder upon rather than spoon feeding a code.
The interpreter does not know how to come back to a line which has already been printer.
Therefore, you would have to draw both sides of the X in one iteration of the loop.
After that you decrease you star count (star--) and draw line # 2.
Repeat till the mid way mark when your stars are 0.
When your code sees that the stars are 0, then start with the same loop but this time star++ in each iteration.
This is repeated till the starting count of the star i.e 4 in my case.
If any problems you can post your code on the site :)

Answer 3

You shall dynamically evaluate the spacing areas in each row. Draw manually desired shape on piece of paper and try to create function, which takes as an argument the row number and returns amount of spaces required in the specific row. For example:
* *
* *
*
* *
* *
The amount of spaces in each row equals:
0 [*] 3 [*]
1 [*] 1 [*]
2 [*]
1 [*] 1 [*]
0 [*] 3 [*]
Note, that inside each row you'll need two loops: first for the initial and middle spaces.

Answer 4

The solution I wrote 2 decades ago (when I was still learning):
Make an line-column array, e.g. char screen[80][25];
Clear it by setting all entries to ' '
"Draw a point" at x,y by setting screen[x][y]='*';
When done, render the whole screen[80][25] by calling cout 2080 times. (2000 times for the characters and and 80 times for endl)
In your case, you know how to draw a \. You can easily adapt this. But with my method, you can then draw a / in the same screen array. And when you're done, you have an overlapping / and \ at the last step: X
I used this method since we had to draw a circle and that's really a lot harder. And yes, nowadays I'd probably use std::vector<std::string> screen but back then screens were really 80x25 :)

Answer 5

#include "stdafx.h"
#include <iostream>
using namespace std;
int main(int argc, char* argv[])
{
int i, d, a=1,b=12,c ,e=1;
for(i = 1; i <= 6; i++)
{
for (d=1; d<i;d++) {cout <<" ";}
cout<<"***";
for (c=a+1; c<b;c++) {cout <<" ";}
{cout<<"***";}
for(b=12-i;b<i;b++)
{cout<<"***";}
cout<<endl;
a++;
}
for( i = 1; i <= 3; i++)
{
for ( d=6; d>i;d--) {cout <<" ";}
cout<<"***";
for (c=0; c<e-1;c++) {cout <<" ";}
{cout<<"***";}
cout<<endl;
e++;
}
return 0;
}

Answer 6

int n=11,i=0,k=0,j=0;
for(i=0;i<n;i++)
{
if(i<(n/2))
{
cout<<endl;
for(j=0;j<i;j++)
{
cout<<" ";
}
cout<<"*";
for(k=n/2;k>i;k--)
{
cout<<" ";
}
cout<<"*";
}
else
{
cout<<endl;
for(k=n-1;k>i;k--)
{
cout<<" ";
}
cout<<"*";
for(j=n/2;j<i;j++)
{
cout<<" ";
}
cout<<"*";
}
}

Answer 7

#include<iostream>
using namespace std;
int main()
{
int i, j;
for(i = 1;i<= 5;i++)
{
for(j = 1;j<= 5;j++)
{
if((i == j)||(j==(5+1)-i))
{
cout << "*";
}
else{
cout << " ";
}
}
cout<< endl;
}
system("pause");
}

Answer 8

#include "stdafx.h"
#include <iostream>
using namespace std;;
int _tmain(int argc, _TCHAR* argv[])
{
int i,k,j;
for (i=1;i<8;i++)
{
for (int k=0;k<i;k++)
{
cout<<" ";
}
cout<<"*";
for (int k=8;k>i;k--)
{
cout<<" ";
}
cout<<"*";
cout<<endl;
}
for (i=1;i<8;i++)
{
for (int k=8;k>i;k--)
{
cout<<" ";
}
cout<<"*";
for (int k=0;k<i;k++)
{
cout<<" ";
}
cout<<" *";
cout<<endl;
}
system("Pause");
return 0;
}