Question

what is the meaning of: i=-n

in here: for(i=-NUM; i<=NUM; i++)

Answer 1

This will make i equal to the negative of NUM. So if NUM = 3, the for loop will loop
between
-3
-2
-1
0
1
2
and finish with i = 3 after the for loop.

Answer 2

how about this one? what is the meaning of this: if( abs(i)+abs(j)<=NUM)

Answer 3

abs means absolute. It basically ignores the minus sign if it exists, so abs(-10) would be 10. abs(i)+abs(j) means take the value that's in i, strip off the minus sign and add it to the value in j without it's minus sign.

<= means less than or equal to, so the whole thing means: if the sum of i and j (ignoring minus signs) is less than or equal to the value in NUM, then...

do the next statement (whatever that is).

Answer 4

can u show me a code that can make a output showing a square with a hole. when i enter 2 number for row and column. for example i enter n(column)=5, m(row)=3:

*****
* *
*****

Answer 5

Here is a piece of code that will print the box of stars.

#include<stdio.h>
int main(int argc,char **argv)
{
int column;
int row;
int x;

if (argc < 3)
{
printf("Not enough arguments\n");
return 1;
}

column = atoi(argv[1]);
row = atoi(argv[2]);

if (row < 2)
{
printf("not enough rows\n");
return 1;
}
printheaderfooter(column);
for(x=1;x < row -1; x++)
{
printcontent(column);
}
printheaderfooter(column);
printf("%d %d", column,row);
}

void printheaderfooter(int length)
{
int i;
for(i=0;i<length;i++)
{
printf("*");
}
printf("\n");
}

void printcontent(int length)
{
int i;
printf("*");
for(i=0;i<length - 2;i++)
{
printf(" ");
}
printf("*\n");
}

Answer 6

wow ?! can u do it using for loop? im not yet familiar with the other loops

Answer 7

can u show me a code. printing a pyramid ? for example, i type 5. this would be the output:

*
**
***
****
*****

using c++

Answer 8

The previous example used for loops in function calls, if you want direct code then you can I can restructure the code to use just the main function.


Pyramid example:
#include <iostream>
using namespace std;
int main()
{
int count;
cin >> count;
for(int i = 1; i <= count ; i++)
{
for (int j = 1; j <= i; j++)
{
cout << "*" ;
}
cout << endl;

}
return 0;
}