1 /*
2  * Exercise 3-5. Write the function itob(n,s,b) that converts the integer n
3  * into a base b character representation in the string s. In particular,
4  * itob(n,s,16) formats n tas a hexadecimal integer in s.
5  */
6 
7 #include <stdio.h>
8 #include <stdlib.h>
9 #include <limits.h>
10 #include <string.h>
11 
12 int pow(int x, int n) {
13 	int val;
14 	for (val = 1; n > 0; n--) {
15 		val *= x;
16 	}
17 	return val;
18 }
19 int itob(int n, char *s, int b) {
20 	char *tr = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
21 	int digits;
22 	for (digits = 0; pow(b,digits) <= n; digits++)
23 		;
24 	for (int i = digits-1; i >= 0; i--) {
25 		int times = (n/pow(b,i));
26 		*s++ = *(tr + times);
27 		n -= times * pow(b,i);
28 	}
29 	*s = '\0';
30 	return 0;
31 }
32 int assert(char *expect, char *actual) {
33 	printf("Expect: %s\n", expect);
34 	printf("Actual: %s\n", actual);
35 	return strcmp(expect, actual);
36 }
37 int main() {
38 	char s[100];
39 	char *t;
40 	sprintf(t, "%d", pow(9,9));
41 	assert("387420489", t);
42 	itob(1208, s, 26);
43 	assert("1KC", s);
44 	itob(5763, s, 16);
45 	assert("1683", s);
46 	itob(194200, s, 8);
47 	assert("573230", s);
48 	itob(2, s, 8);
49 	assert("2", s);
50 	itob(8, s, 2);
51 	assert("1000", s);
52 	itob(93, s, 2);
53 	assert("1011101", s);
54 	/*
55 	 * overflow issues
56 	sprintf(t, "%x", INT_MAX);
57 	itob(INT_MAX, s, 16);
58 	assert(t, s);
59 	*/
60 }
61 
62 /*
63  * thought process:
64  * 1208/26
65  * 46.46153846153846153846
66  * too big, repeat division
67  * 1208/26/26
68  * 1.78698224852071005917
69  * base 26: 0100
70  * 1 + modulo
71  * 26*26 = 676
72  * 1208 - 676 = 532
73  * 532 / 26 = 20
74  * lookup 20 (char 9abcdefghijk)
75  * base 26: 01k0 + modulo
76  * 532 - 20*26 = 12
77  * lookup 20 (char c)
78  * base 26: 01kc
79  * 5763/16 = 360.18
80  * too big, repeat division
81  * 5763/16/16 = 22.511
82  * too big, repeat division
83  * 5763/16/16/16 = 1.406
84  * base 16: 1 + modulo
85  * 5763-1*16*16*16 = 1667
86  * 1667/16/16 = 6.5117
87  * 1667-6*16*16 = 131
88  * base 16: 16 + modulo
89  * 131/16 = 8.1875
90  * 131-8*16 = 3
91  * base 16: 163
92  * 194200/8/8/8/8/8 = 5.926
93  * 194200 - 5*8^5 = 30360
94  * 30360/8/8/8/8 = 7.41
95  * 30360 - 7*8^4 = 1688
96  * 1688/8/8/8 = 3.2968
97  * 1688 - 3*8^3 = 152
98  * 152/8/8 = 2.375
99  * 152 - 2*8^2 = 24
100  * 24/8 = 3
101  * base 8: 573230
102  */
103 
104 /*
105 itob(INT_MIN, s, 2);
106 printf("Expect: %s\n", "");
107 printf("Actual: %s\n", s);
108 itob(-5, s, 8);
109 printf("Expect: %s\n", "");
110 printf("Actual: %s\n", s);
111 */
112