Line pattern in a diamond in C - c

Question

I'm a beginner in C, I'm facing a problem in order to implement a diamond, I'm following guided exercises from a book but I got stuck, I have to implement the pattern:
According to the exercise, they suggest me to implement the pattern using the operator %
I coded the structure but when I want to implement the pattern #.o.# I just destroy everything
#include<stdio.h>
int main(){
int n, space;
printf("Number of sides: ");
scanf("%d", &n);
space = n - 1;
for (int i = 0; i < n; i++){
for (int j = 0;j < space; j++){
printf(" ");
}
// patern #.o should be in this loop
for (int j = 0;j <= i*2; j++){
printf("#");
}
printf("\n");
space--;
}
space = 0;
for (int i = n; i >= 0; i--){
for (int j = 0; j < space; j++){
printf(" ");
}
// patern #.o should be in this loop
for (int j = 0;j < (i*2)-1;j++){
printf("#");
}
printf("\n");
space++;
}
}
I really appreciate it if somebody could give me a hint or help with this, I got stuck here for the last 2 weeks.
Thank you.

Answer 1

This is what you need
#include<stdio.h>
int main(){
int n, space,i,j;
char pattern[]={'#','.','o','.'};
int position;
printf("Number of sides: ");
scanf("%d", &n);
space = n - 1;
for (i = 0; i < n; i++){
position=0;
for (j = 0;j < space; j++){
printf(" ");
}
// patern #.o should be in this loop
for (j = 0;j <= i; j++,position++)
printf("%c",pattern[position%sizeof(pattern)]);
position-=2;
for(;j<=2*i;j++,position--)
printf("%c",pattern[position%sizeof(pattern)]);
printf("\n");
space--;
}
}
The output for some cases:
./test
Number of sides: 10
#
#.#
#.o.#
#.o.o.#
#.o.#.o.#
#.o.#.#.o.#
#.o.#.o.#.o.#
#.o.#.o.o.#.o.#
#.o.#.o.#.o.#.o.#
#.o.#.o.#.#.o.#.o.#
./test
Number of sides: 5
#
#.#
#.o.#
#.o.o.#
#.o.#.o.#
./test
Number of sides: 2
#
#.#
For a diamond:
#include<stdio.h>
int main(){
int n, space,i,j;
char pattern[]={'#','.','o','.'};
int position;
printf("Number of sides: ");
scanf("%d", &n);
space = n - 1;
for (i = 0; i < n; i++){
position=0;
for (j = 0;j < space; j++){
printf(" ");
}
for (j = 0;j <= i; j++,position++)
printf("%c",pattern[position%sizeof(pattern)]);
position-=2;
for(;j<=2*i;j++,position--)
printf("%c",pattern[position%sizeof(pattern)]);
printf("\n");
space--;
}
space = 0;
for (int i = n-1; i >= 0; i--){
for (int j = 0; j < space; j++){
printf(" ");
}
position=0;
for (j = 0;j <= i; j++,position++)
printf("%c",pattern[position%sizeof(pattern)]);
position-=2;
for(;j<=2*i;j++,position--)
printf("%c",pattern[position%sizeof(pattern)]);
printf("\n");
space++;
}
}
And the output
./test
Number of sides: 10
#
#.#
#.o.#
#.o.o.#
#.o.#.o.#
#.o.#.#.o.#
#.o.#.o.#.o.#
#.o.#.o.o.#.o.#
#.o.#.o.#.o.#.o.#
#.o.#.o.#.#.o.#.o.#
#.o.#.o.#.#.o.#.o.#
#.o.#.o.#.o.#.o.#
#.o.#.o.o.#.o.#
#.o.#.o.#.o.#
#.o.#.#.o.#
#.o.#.o.#
#.o.o.#
#.o.#
#.#
#

Answer 2

you have to just change printf("#") to printf("%c",pattern[j%4])
Here is a code:
int main(){
int n, space;
char pattern[4] = {'#','.','o','.'};
printf("Number of sides: ");
scanf("%d", &n);
space = n - 1;
for (int i = 0; i < n; i++){
for (int j = 0;j < space; j++){
printf(" ");
}
// patern #.o should be in this loop
for (int j = 0;j <= i*2; j++){
printf("%c",pattern[j%4]);
}
printf("\n");
space--;
}
space = 0;
for (int i = n; i >= 0; i--){
for (int j = 0; j < space; j++){
printf(" ");
}
// patern #.o should be in this loop
for (int j = 0;j < (i*2)-1;j++){
printf("%c",pattern[j%4]);
}
printf("\n");
space++;
}
}
The reason why you "destroy everything" is probably because you tried to do it with printf(pattern[j%4]). printf as first parameter wants string (address in memory when string is stored) when you pass pattern[j%4], it thinks that for example '#' (in ASCII 64) is adress in memory, and want to read from it, but it is not valid memory adress so operating system kill your program and it "get's destroyed"