[Algorithms, C] long2char: 自己用C写来玩玩的整形转化为字符串函数 (递归和迭代两种实现): void long2char(long from, char *to, int n);

void long2char(long from, char *to, int n); // 递归和迭代两种实现

 

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/* long2char.c * Author: dooit.lee@gmail.com * Date: Sun Jun 19, 2011 */ #include <stdio.h> #include <stdlib.h> /* * from: int, to: char*, `n' means n digits of `from' with to[n + 2] */ void long2char(long from, char *to, int n) { int digit; if (from < 0) { *to = '-'; to++; from = -1 * from; } #if 0 // Recursion digit = from % 10; if ((from /= 10) != 0) { long2char(from, to, n - 1); } to[n - 1] = '0' + (digit - 0); #else // Iteration do { digit = from % 10; to[--n] = '0' + (digit - 0); } while ((from /= 10) != 0); #endif } int main(int argc, char *argv[]) { long from; int n, i; char *to, *endptr; if (argc != 2 || !atoi(argv[1])) { bad: fprintf(stderr, "Usage: %s <integer-number>/n", argv[0]); exit(1); } else { from = strtol(argv[1], &endptr, 10); if (argv[1] == endptr) goto bad; n = 1; i = from; while ((i /= 10) != 0) ++n; // include a '-' and '/0' charachter to = calloc(n + 2, sizeof(char)); if (to != NULL) { long2char(from, to, n); printf("%s/n", to); free(to); } else { fprintf(stderr, "%s: Cannot allocate memory/n", argv[0]); } } return 0; }