練習3.4

編程：編寫itoa(int a, char s[])函數，將數字轉換成字符串並保存在s中。

 

代碼：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <limits.h>

void itoa(int n, char s[]);
void reverse(char *s);
void itoa1(int n, char s[]);
int main(int argc, char *argv[])
{
    int n = 0;
//    n = INT_MAX;

    char s[20] = "";
    itoa(n, s);
    printf("%s\n", s);

    char s2[20] = "";
    itoa1(n, s2);
    printf("%s\n", s2);



    n = INT_MIN;
    itoa(n, s);
    printf("%s\n", s);

    itoa1(n, s2);
    printf("%s\n", s2);


    return 0;
}
void itoa1(int n, char s[])
{
    char neg = 'y';
    if (n >= 0) {
        neg = 'n';//不是負數
    }
    if (n == 0) {
        s[0] = '0';
        s[1] = '\0';
        return ;
    }
    int i = 0;
    while (n != 0) {
        int k = n % 10;
        if (k < 0) k = -k;
        n /= 10;
        s[i++] = '0'+k;
    }
    if (neg == 'y')
        s[i++] = '-';
    s[i] = '\0';
    reverse(s);
}
void reverse(char *s)
{
    int i = 0;
    int j = strlen(s)-1;
    while (i < j) {
        char item = s[j];
        s[j] = s[i];
        s[i] = item;
        ++i,--j;
    }
}

void itoa(int n, char s[])
{
    int i, sign;
    if ((sign = n) < 0)
        n = -n;
        //就是因爲這條語句，不能處理最小的負數
    i = 0;
    do {
        s[i++] = n % 10 + '0';
    } while ((n /= 10) > 0);
    if (sign < 0)
        s[i++] = '-';
    s[i] = '\0';
    reverse(s);
}

這些都沒有什麼說的。。。