大家帮帮忙，此函数的非递归实现该怎么做啊？
void printn(uint n, uint b){
static char *ntab = "0123456789ABCDEF";
uint a, m;
if (a = n / b){
printn(a, b);
}
m = n % b;
myputchar( ntab[m] );
}


[解决办法]
你这不是尾递归，非尾递归而要接触递归一般都伴随代码量大幅增加，所以楼主还是别了
[解决办法]

C/C++ code

void printn(int n, int b){    static char *ntab = "0123456789ABCDEF";    int nrst = 1;    static unsigned short usrst[20] = {0};    while (n != 0)    {        usrst[nrst++] = n % b;        n = n / b;    }    while (--nrst)    {        //printf("%c",ntab[usrst[nrst]]);       myputchar(ntab[usrst[nrst]]);    }}
[解决办法]
晕，我测试的是9，所以到这和正着的一个样，所以没察觉。。。不好意思哈
C/C++ code
void printn(uint n, uint b){    static char *ntab = "0123456789ABCDEF";    static char pcPut[128] = {0};    uint a, m=0,i=0;    while(a = n / b){        pcPut[m] = n % b;        n = a;        m++;    }    pcPut[m] = n % b;    for (i = m;i >= 0;i++)    {        myputchar(ntab[pcPut[i]]);    }}