## alfers101 3 years ago what is the meaning of: i=-n in here: for(i=-NUM; i<=NUM; i++)

1. ecdown

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.

2. alfers101

3. Toolan

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).

4. alfers101

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: ***** * * *****

5. ecdown

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"); }

6. alfers101

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

7. alfers101

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

8. ecdown

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; }