Completed day 2 part 1 and 2

2015_day_2/Makefile

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+compile:
+	gcc day_2.c -Wall -Werror=switch -o day_2
+
+run:
+	./day_2

2015_day_2/README

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+Note: My goal here is to learn C better, not just complete the challenge, so in this case I'm learning structs a little better, and trying a 2d array.
+Completed 22/3/2024
+
+
+Prompt:
+--- Day 2: I Was Told There Would Be No Math ---
+The elves are running low on wrapping paper, and so they need to submit an order for more. They have a list of the dimensions (length l, width w, and height h) of each present, and only want to order exactly as much as they need.
+
+Fortunately, every present is a box (a perfect right rectangular prism), which makes calculating the required wrapping paper for each gift a little easier: find the surface area of the box, which is 2*l*w + 2*w*h + 2*h*l. The elves also need a little extra paper for each present: the area of the smallest side.
+
+For example:
+
+A present with dimensions 2x3x4 requires 2*6 + 2*12 + 2*8 = 52 square feet of wrapping paper plus 6 square feet of slack, for a total of 58 square feet.
+A present with dimensions 1x1x10 requires 2*1 + 2*10 + 2*10 = 42 square feet of wrapping paper plus 1 square foot of slack, for a total of 43 square feet.
+All numbers in the elves' list are in feet. How many total square feet of wrapping paper should they order?
+
+
+
+
+Your puzzle answer was 1588178.
+
+The first half of this puzzle is complete! It provides one gold star: *
+--- Part Two ---
+
+The elves are also running low on ribbon. Ribbon is all the same width, so they only have to worry about the length they need to order, which they would again like to be exact.
+
+The ribbon required to wrap a present is the shortest distance around its sides, or the smallest perimeter of any one face. Each present also requires a bow made out of ribbon as well; the feet of ribbon required for the perfect bow is equal to the cubic feet of volume of the present. Don't ask how they tie the bow, though; they'll never tell.
+
+For example:
+
+    A present with dimensions 2x3x4 requires 2+2+3+3 = 10 feet of ribbon to wrap the present plus 2*3*4 = 24 feet of ribbon for the bow, for a total of 34 feet.
+    A present with dimensions 1x1x10 requires 1+1+1+1 = 4 feet of ribbon to wrap the present plus 1*1*10 = 10 feet of ribbon for the bow, for a total of 14 feet.
+
+How many total feet of ribbon should they order?
+ Total ribbon needed: 3783758 

2015_day_2/day_2.c

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+#include<stdio.h>
+#include<stdlib.h>
+#include<string.h>
+FILE *fptr;
+
+struct dim_struct{
+    int dimensions[3];
+    int surfaces[3];
+    int smallest_side;
+    int smallest_side_id;
+    int smallest_side_perimeter;
+    int box_volume;
+    int total_ribbon_for_item;
+    int total_paper_for_item;
+};// Fun fact: C expects a semicolon after the struct definition
+
+int main(){
+struct dim_struct myArray[1005];
+fptr = fopen("2015_day_2_input.txt", "r");  
+if(fptr == NULL) {
+  printf("Not able to open the file.");
+  return 1;
+}
+printf("hello\n");
+long int line_count = 0;
+int dim = 0;
+int matrix[3][2] = { {0,1}, {1,2}, {2,0} };
+int i;
+int dim1, dim2;
+char buffer[266];
+while (fgets(buffer, sizeof(buffer), fptr)) {
+    printf("%s\n",buffer);
+    char delimiter = 'x';
+    char *token, *saveptr;
+    line_count++;
+    for (token = strtok_r(buffer, &delimiter, &saveptr); token != NULL; token = strtok_r(NULL, &delimiter, &saveptr)) {
+        printf("%s\n", token); // or process token
+        myArray[line_count].dimensions[dim] = atoi(token);
+        printf("\nwow\n%d\nwow\n",myArray[line_count].dimensions[dim]);
+        dim++;
+    }
+    memset(buffer, 0, sizeof(buffer)); // clear the buffer to zero
+    dim = 0;
+}
+
+printf("\n length of myArray = %lu\n", sizeof(myArray));
+for (i=0; i<line_count+1; i++){
+    myArray[i].surfaces[0] = myArray[i].dimensions[0] * myArray[i].dimensions[1];
+    myArray[i].surfaces[1] = myArray[i].dimensions[1] * myArray[i].dimensions[2];
+    myArray[i].surfaces[2] = myArray[i].dimensions[2] * myArray[i].dimensions[0];
+    if ((myArray[i].surfaces[0] <= myArray[i].surfaces[1]) && (myArray[i].surfaces[0] <= myArray[i].surfaces[2])){
+        myArray[i].smallest_side = myArray[i].surfaces[0];
+        myArray[i].smallest_side_id = 0;
+    }
+    else if ((myArray[i].surfaces[1] <= myArray[i].surfaces[0]) && (myArray[i].surfaces[1] <= myArray[i].surfaces[2])){
+        myArray[i].smallest_side = myArray[i].surfaces[1];
+        myArray[i].smallest_side_id = 1;
+    }
+    else if ((myArray[i].surfaces[2] <= myArray[i].surfaces[0]) && (myArray[i].surfaces[2] <= myArray[i].surfaces[1])){
+        myArray[i].smallest_side = myArray[i].surfaces[2];
+        myArray[i].smallest_side_id = 2;
+    }
+    else {
+        printf("ERROR!");
+        exit(1);
+    }
+  
+    myArray[i].total_paper_for_item = ((2 * (myArray[i].surfaces[0] + myArray[i].surfaces[1] + myArray[i].surfaces[2])) + myArray[i].smallest_side);
+    printf("\ntotal paper for item = %d\n", myArray[i].total_paper_for_item);
+    printf("surface 0 is %d, %d times %d\n", myArray[i].surfaces[0], myArray[i].dimensions[0] , myArray[i].dimensions[1]);
+    printf("surface 1 is %d, %d times %d\n", myArray[i].surfaces[1], myArray[i].dimensions[1] , myArray[i].dimensions[2]);
+    printf("surface 2 is %d, %d times %d\n", myArray[i].surfaces[2], myArray[i].dimensions[2] , myArray[i].dimensions[0]);
+    printf("The smallest side of item %d is %d\n",i ,myArray[i].smallest_side);
+  
+
+    myArray[i].smallest_side_perimeter = (2 * myArray[i].dimensions[matrix[myArray[i].smallest_side_id][0]]) + (2 * myArray[i].dimensions[matrix[myArray[i].smallest_side_id][1]]); //don't love this, but it seems to be a fairly simple way to do it
+    printf("\n perimeter of smallest side = %d\n", myArray[i].smallest_side_perimeter);
+    myArray[i].box_volume = myArray[i].dimensions[0] * (myArray[i].dimensions[1] * myArray[i].dimensions[2]);
+    myArray[i].total_ribbon_for_item = myArray[i].smallest_side_perimeter + myArray[i].box_volume;
+
+}
+for (i=0; i<line_count+1; i++){
+    dim1 = dim1 + myArray[i].total_paper_for_item;
+    dim2 = dim2 + myArray[i].total_ribbon_for_item;
+
+}
+
+fclose(fptr);
+printf("\n\n Total paper needed: %d \n\n", dim1);
+printf("\n\n Total ribbon needed: %d \n\n", dim2);
+
+return 0;
+}
+//https://www.geeksforgeeks.org/how-to-create-an-array-of-structs-in-c/