anonymous
  • anonymous
what is the meaning of: i=-n in here: for(i=-NUM; i<=NUM; i++)
Computer Science
schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
how about this one? what is the meaning of this: if( abs(i)+abs(j)<=NUM)
anonymous
  • anonymous
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).

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anonymous
  • anonymous
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: ***** * * *****
anonymous
  • anonymous
Here is a piece of code that will print the box of stars. #include 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
anonymous
  • anonymous
wow ?! can u do it using for loop? im not yet familiar with the other loops
anonymous
  • anonymous
can u show me a code. printing a pyramid ? for example, i type 5. this would be the output: * ** *** **** ***** using c++
anonymous
  • anonymous
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 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; }

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