How do I properly nest these loops in C? - c

Question

I need to make a program that prints this:
.......#
......##
.....###
....####
...#####
..######
.#######
########
Basically, if the n is 8. it will print 7 dots and 1 hash***. then in the next row it will print 7 dots and 2 hashes, then 6 dots and 3 hashes until there are 0 dots and 8 hashes.
So far my code looks like this:
int main(void)
{
int G = 1;
int n = 8;
int m = 0;
int k = 0;
int Z = 1;
for (m = n; m > 0; m --)
{
for (k = n - 1; k >= Z; k --)
{
printf(".");
}
printf("\n");
Z = Z + 1;
}
for (int i = 0; i < n; i ++)
{
for (int j = 0; j < G; j++)
{
printf("#");
}
printf("\n");
G = G + 1;
}
}
But the result of this comes out as this:
.......
......
.....
....
...
..
.
#
##
###
####
#####
######
#######
########

Answer 1

Since this is obviously homework or some self studies, I'll write how you need to think here.
The first problem is to be able to print ONE row given total number of rows and current row.
So let's start with total number or rows. I will use more descriptive variable names, which you also should.
int total = 8;
Now we need to print one row, given the current row. Let's assume that the first row has number 1. Let's start with the dots.
for(int i=0; i<(total-row); i++)
printf(".");
Then we continue with the hashes:
for(int i=0; i<row; i++)
printf("#");
And now we just need to do this total times and add a newline after each:
for(int row=1; row<total; row++) {
for(int i=0; i<(total-row); i++)
printf(".");
for(int i=0; i<row; i++)
printf("#");
printf("\n");
}
Please note that it is very common with obo (off by one) errors when doing things like this. We could let 0 be the first row instead of one, but then we need to change a few other things. You said you wanted 1 hash and 7 dots on first line and 8 hash and 0 dots on last line, which seems a bit odd, but just change row<total to row<total+1 and you get that.

Answer 2

You'll rather want something that does the printing from within the same loop. Here's some pseudo code of a simple way to write this algorithm:
for(int dots=n-1; dots>DOTS_MIN; dots--)
{
for(int i=0; i < dots; i++)
print dot
for(int i=dots; i < n; i++)
print hash
printf("\n");
}

Answer 3

First, start from naming, it's a challenging task to find out what Z stands for.
So we have
int totalRows = 4;
int totalColumns = 8;
and we should print out totalRows of rows:
for (int row = 1; row <= totalRows; ++row) {
}
from 2nd row on we should add a new line:
for (int row = 1; row <= totalRows; ++row) {
/* Start a new line from the 2nd row on */
if (row > 1)
printf("\n");
}
Time to print dots, we have totalColumns - row of them:
for (int i = 1; i <= totalColumns - row; ++i)
printf(".");
Followed by row # symbols:
for (int i = 1; i <= row; ++i)
printf("#");
Combining these chunks all together:
int main() {
int totalRows = 4;
int totalColumns = 8;
for (int row = 1; row <= totalRows; ++row) {
/* Start a new line from the 2nd row on */
if (row > 1)
printf("\n");
/* Printing dots */
for (int i = 1; i <= totalColumns - row; ++i)
printf(".");
/* Printing # */
for (int i = 1; i <= row; ++i)
printf("#");
}
return 0;
}