How this program executes this result!?

Question

#include "stdafx.h"
#include "iostream"
using namespace std;
void main()
{
int x = 0;
int j;
for ( x=1; x<=5 ; x++){
for (j =1; j<=x; j++)
cout << x << endl;
}
}
Results :- 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5.

Answer 1

There's nothing mind-blowing about it, especially when it uses void return type for main.

http://www.stroustrup.com/bs_faq2.html#void-main

Now, back on topic. The outer loop runs for values from 1 to 5. These values are represented by x, within the loop. It uses a nested loop to loop for x amount of times, each printing the value of x. Hence, if the value of x is 3, then 3 will be printed for 3 times. If the value of x is 4, then 4 will be printed 4 times. Similarly, if the value of x is n, then n will be printed for n times.

Answer 2

@cimraan

#include "stdafx.h"
#include "iostream"
using namespace std;
void main()
{
int x = 0;
int j;
for ( x=1; x<=5 ; x++){. //line 8
for (j =1; j<=x; j++). //line 9
cout << x << endl;
}
}
Results :- 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5

there are 2 loops(lets call ,1st for line 8,2nd for line 9)
the1st one takes the value from 1 to 5
and 2nd takes the value from 1 to value of x from 1st loop

hence
when x passes as 1 in 1st loop
2nd loop takes x value from 1 to 1
output 1

when x passes as 2 in 1st loop
2nd loop takes x value from 1 to 2 means 1,2 (printing 2 times the x value)
output 1 2 2

when x passes as 3 in 1st loop
2nd loop takes x value from 1 to 3 means 123
printing 3 times the x value
output 1 2 2 3 3 3

similarly for 4 and 5

....................

Answer 3

still didn't get it😥 but thanks a lot @Rei