solution using arbitrary length decimal number addition using array allocation

project_euler/Problem 13/sol1.c

@@ -0,0 +1,136 @@
+#include <stdio.h>
+#include <stdint.h>
+#include <stdlib.h>
+#include <string.h>
+
+/* Function to read the number from a file and store it in array.
+    index 0 of output buffer => units place
+    index 1 of output buffer => tens place and so on
+    i.e., index i => 10^i th place
+ */
+int get_number(FILE *fp, char *buffer, uint8_t *out_int)
+{
+    long l = fscanf(fp, "%s\n", buffer);
+    if (!l) 
+    {
+        perror("Error reading line."); 
+        return -1;
+    }
+    // printf("Number: %s\t length: %ld, %ld\n", buffer, strlen(buffer), l);
+    
+    long L = strlen(buffer);
+
+    for (int i = 0 ; i < L; i++)
+        if (buffer[i] < 0x30 || buffer[i] > 0x39)
+        {
+            perror("found inavlid character in the number!");
+            return -1;
+        } else
+            out_int[L-i-1] = buffer[i] - 0x30;
+
+    return 0;
+}
+
+/**
+ * Function to add arbitraty length decimal integers stored in an array.
+ * a + b = c = new b
+ **/
+int add_numbers(uint8_t *a, uint8_t *b, uint8_t N)
+{
+    int carry = 0;
+    uint8_t *c = b; /* accumulate the result in the array 'b' */
+
+    for (int i = 0; i < N; i++)
+    {
+        // printf("\t%d + %d + %d ", a[i], b[i], carry);
+        c[i] = carry + a[i] + b[i];
+        if (c[i] > 9)   /* check for carry */
+        {
+            carry = 1;
+            c[i] -= 10;
+        } else
+            carry = 0;
+        // printf("= %d, %d\n", carry, c[i]);
+    }
+
+    for (int i = N; i < N+10; i++)
+    {
+        if(carry == 0)
+            break;
+        // printf("\t0 + %d + %d ", b[i], carry);
+        c[i] = carry + c[i];
+        if (c[i] > 9)
+        {
+            carry = 1;
+            c[i] -= 10;
+        } else
+            carry = 0;
+        // printf("= %d, %d\n", carry, c[i]);
+    }
+    return 0;
+}
+
+int print_number(uint8_t *number, uint8_t N, int8_t num_digits_to_print)
+{
+    uint8_t start_pos = N - 1;
+    uint8_t end_pos;
+
+    /* skip all initial zeros */
+    while (number[start_pos] == 0)
+        start_pos --;
+
+    /* if end_pos < 0, print all digits */
+    if (num_digits_to_print < 0)
+        end_pos = 0;
+    else if (num_digits_to_print <= start_pos)
+        end_pos = start_pos - num_digits_to_print + 1;
+    else
+    {
+        fprintf(stderr, "invalid number of digits argumet!\n");
+        return -1;
+    }
+    
+    for (int i = start_pos; i >= end_pos ; i--)
+        putchar(number[i] + 0x30);
+    
+    putchar('\n');
+
+    return 0;
+}
+
+
+int main(void)
+{
+    const char N = 50, N2 = N+10;          /* length of numbers */
+    char txt_buffer[N+5];       /* temporary buffer */
+    uint8_t number[N];          /* array to store digits of a large number */
+    uint8_t sum[N2];          /* array to store the sum of the large numbers. For 
+                                safety, we make it twice the length of a number. */
+
+    memset(sum, 0, sizeof(sum));    /* initialize sum array with 0 */
+
+    FILE *fp = fopen("num.txt", "rt");  /* open text file to read */
+    if(!fp)
+    {
+        perror("Unable to open file 'num.txt'.");
+        return -1;
+    }
+
+    int count = 0;
+    get_number(fp, txt_buffer, sum);    /* 0 + = first_number = first_number */
+    do {
+        count ++;
+        if (get_number(fp, txt_buffer, number) != 0)
+            break;
+        add_numbers(number, sum, N);
+    } while (!feof(fp));
+
+    printf("\nSum   : ");
+    print_number(sum, N2, -1);
+
+    printf("first 10 digits: \t");
+    print_number(sum, N2, 10);
+
+    fclose(fp);     /* close file */
+    return 0;
+}