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assignment8.txt

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+Caleb Gershengorn
+
+1:
+It will print a,a (the zeroth term in the array which prints twice because it is twice not equal to 3)
+I think it would then return 1 and end the function, but that seems completely wrong, but it makes sense since the code is at one point not equal to 3, thus it would return 1 and end.
+
+2:
+Based on the logic from the first answer, none of the code from underneath the return 1 runs because the return statement has terminated the code.
+
+3:
+Assuming that I am mistaken about the return ending the code, the lines with memory errors would be lines 27-28 and 19-20
+
+4:
+I honestly do not know, but I would guess that you could make the q in line 28 a dereference operator tomake sure that the memory is called and thus not lost.

sort.c

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+#include <stdio.h>
+#include <time.h>
+#include <stdlib.h>
+//I used the sorting algorithm pseudocode from here: https://en.wikipedia.org/wiki/Insertion_sort
+int sorting(length){
+	srand(time(NULL));
+	int i,j,l,m,p,q,c;
+	int* arr1=(int*)malloc(length*4);
+	for (i=0;i<length;i++)
+	{
+		int num = rand()%9+1;	
+		arr1[i]=num;
+	}	
+	printf("\n\nThe first array is: \n");
+	for (j=0;j<length;j++)
+	{
+		printf("%d ",arr1[j]);
+	}
+	int*arr2=(int*)malloc(length*4);
+	for (q=0;q<length;q++)
+	{
+	arr2[q]=arr1[q];
+	}
+	for (l=0;l<length;l++)
+	{
+		int x =arr2[l];
+		int k = l-1;
+		while (k>=0&&arr2[k] > x)
+		{
+			arr2[k+1]=arr2[k];
+			k=k-1;
+		}	
+		arr2[k+1]=x;
+	}	
+	printf("\n\nThe second array is: \n");
+	for (m=0;m<length;m++)
+	{
+		printf("%d ",arr2[m]);
+	}
+	int*arr3=(int*)malloc(length*4);
+	for (p=0;p<length;p++)
+	{
+		arr3[p]=arr2[(length-1)-p];
+	}
+	printf("\n\nThe third array is: \n");
+	for (c=0;c<length;c++)
+	{
+	printf("%d ",arr3[c]);
+	}
+	printf("\n");
+	return 0;
+}
+
+int main(){
+	int length;
+	printf("Enter the length of array you want: \n");
+	scanf("%d",&length);
+	sorting(length);
+}