my_itob

K&R第三章习题，把十进制数转化成任意进制（2、8、16）

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/* * Write the function itob(n,s,b) that converts the integer n into a * base b character representation in the string s. In particular, * itob(n,s,16) formats n as a hexadecimal integer in s. */#include <stdio.h>#include <string.h>void reverse(char s[], int n) {int i = 0, j = n;while (i < j) {int temp = s[i];s[i] = s[j];s[j] = temp;i++;j--;}}/*void my_itoa(int n, char s[]) {int sign = (n >= 0) ? 1 : -1;n = (n >= 0) ? n : -n;int i = 0;while(n) {s[i++] = n % 10 + '0';n /= 10;}if(sign == -1) {s[i++] = '-';}s[i] = '\0';reverse(s, strlen(s)-1);}*/void my_itob(int n, char s[], int step) {int sign = (n >= 0) ? 1 : -1;n = (n >= 0) ? n : -n;char *symbols = "0123456789ABCDEF";int i = 0;while(n) {//s[i++] = n % step + '0'; can not deal with hexs[i++] = *(symbols + n % step);n /= step;}if(sign == -1) {s[i++] = '-';}s[i] = '\0';reverse(s, strlen(s)-1);}int main(void){    char s[16];    my_itob(-10, s, 16);    printf("%s\n", s);    return 0;}