Printing * pattern in C++ - c++

Question

I am trying to print a pattern like this
*******
* *
* *
* *
* *
* *
*******
In this it should look like an empty box.
but somehow I am not getting even closer
I coded this so far
#include <iostream>
using namespace std;
int main( int argc, char ** argv ) {
for(int i=1;i<=7;i++)
{
for(int j=1;j<=7;j++)
{
if(j==1||j==7)
printf("*");
else printf(" ");
}
printf("\n");
}
return 0;
}
and my output is
* *
* *
* *
* *
* *
* *
* *
it will be good to have for loop only

Answer 1

if(j==1||j==7)
printf("*");
else printf(" ");
This logic works for all rows except first and last one. So you have to consider row value and make special check for first and last rows. These two do not have spaces.
[Assuming it's a homework, I'm giving just a hint. You almost have done it, above hint should be enough to get this working.]

Answer 2

Your if condition simply needs to be:
if (i==1 || i==7 || j==1 || j==7)
That is, you need to check whether you're on either the first or last rows as well as either the first or last columns, and then print a *.

Answer 3

You are very close. The problem is in this line:
if(j==1||j==7)
Change it so that it also takes into account the top and bottom rows.
Good luck

Answer 4

You need to behave differently during first and last row:
int W = 7, H = 7;
for(int i=0;i<=H;i++)
{
for(int j=0;j<=W;j++)
{
// if row is the first or row is the last or column is the first of column is the last
if (i == 0 || i == H-1 || j == 0 || j == W-1)
printf("*");
else printf(" ");
}
printf("\n");
}

Answer 5

This function will work fine:
#include <iostream>
#include <string>
void printBox(int width, int height) {
if (width < 2 or height < 1) return;
for (int i = 1; i <= height; i++) {
if (i == 1 or i == height) {
std::string o(width, '*');
std::cout << o << std::endl;
} else {
std::string o(width-2, ' ');
o = '*' + o + '*';
std::cout << o << std::endl;
}
}
}
It can be used as:
printBox(2, 2);
which prints:
**
**
Or as:
printBox(6, 4);
which prints:
******
* *
* *
******

Answer 6

Since I'm not expert in programming, I came up with this simple code:
#include <stdio.h>
int main(void)
{
int i,j,k,n;
printf("Enter no of rows : \n");
scanf("%d", &n);
for(i=0; i<n; i++)
{
if(i == 0 || i == (n-1))
{
for(j=0; j <n-1; j++)
printf("*");
}
else
{
for(k=0; k<n; k++)
{
if (k == 0 || k == (n-2))
printf("*");
else
printf(" ");
}
}
printf("\n");
}
return 0;
}

Answer 7

for(int j=1;j<=7;j++)
{
if(j==1||j==7)
printf("*******\n");
else
printf("* *\n");
}
printf("\n");

Answer 8

You are treating all lines equally and ignoring the fact that the first and last line must be handled differently. You need something like
if (i == 1 || i == 7)
{
for (j=1;j<7;j++) printf("*");
printf("\n");
}
else { /* your routine */ }

Answer 9

This is what your code should look like:
#include <iostream>
using namespace std;
int main( int argc, char ** argv ) {
for(int i=1;i<=7;i++)
{
for(int j=1;j<=7;j++)
{
if(j == 1 || j == 7)
printf("*");
else if (i == 1 || i == 7 ) //added this check
printf ("*");
else printf(" ");
}
printf("\n");
}
return 0;
}
Live example

Answer 10

You'll need a nested loop. Because you've two options in between their, you'll need two nested loops. One for spaces and one for the filling '*' in iteration 1 and 7 (0 and 6).
Print the line 1 and 7 with a '*' filling the boundary stars.
Using
if (i == 0 || i == 7) // in-case you're initializing i with 0
// loop for printf ("*");

Answer 11

As others said, you need to handle first and last row. One way to build each line is to make use of one of string constructors, so you don't need to write the inner loop.
#include <string>
#include <iostream>
using namespace std;
int main()
{
int height = 7, width = 7;
for( int i=0; i<height; i++ )
{
char c = (i == 0 || i == height-1) ? '*' : ' ';
string line(width, c);
line[0] = line[width-1] = '*';
cout << line << endl;
}
return 0;
}

Answer 12

#include <iostream>
#include<iomanip>
using namespace std;
main()
{
int Width;
char MyChar;
int LCV; //Loop control
int LCVC, LCVR; //Loop control for LCVC=Columns LCVR=Rows
cout<<"\nEnter Width: "; cin>>Width;
cout<<"\nEnter a Character: "; cin>>MyChar;
for (LCV=1;LCV<=Width;LCV++)
{cout<<MyChar;}
cout<<endl;
for(LCVC=1;LCVC<=Width;LCVC++)
{ cout<<MyChar<<setw(Width-1)<<MyChar<<endl;
}
for (LCV=1;LCV<=Width;LCV++)
{cout<<MyChar;}
cout<<endl;
system("pause");
}
/*
Enter Width: 6
Enter a Character: #
######
# #
# #
# #
# #
# #
# #
######
Press any key to continue . . .
*/

Answer 13

Print this pattern
enter a number=each number for example 23517
**
*
Then
**
*
*****
*
*******
then
** *
* *
***** *
and
*
*
*
*
*

Answer 14

I have come up with a simple solution. You can keep it very simple rather than using char array pointers that you used.
#include <iostream>
using namespace std;
int main() {
int rows = 7;
for(int i = 1; i<=rows;i++){
if(i == 1 || i == rows){
cout<<"*******"<<endl;
}else{
cout<<"* *"<<endl;
}
}
return 0;
} // This code outputs 7 rows with 7 characters in 1st and 7 line each and 1 each character at the start and last position of the remaining rows
But if you want a custom number of rows, you can try out the following code
#include <iostream>
using namespace std;
int main() {
int rows;
cout<<"Enter the number of rows: ";
cin>>rows; // gets input rows from the user
for(int i = 1; i<=rows;i++){
if(i == 1 || i == rows){
cout<<"*******"<<endl;
}else{
cout<<"* *"<<endl;
}
}
return 0;
}

Answer 15

you can add:
`if(i==0 || i==6)
{
cout<<"*";
}`