int i = 12345;char m[20];char* p = m, *q = m;while (i > 0) { *(p++) = '0'+i%10; i /= 10;}*p = 0;p--;while (q < p) {char t = *p; *(p--) = *q; *(q++) = t;}printf("%s", m);
这个题目是用循环好做，用递归还有点烦，还好做出来了：include "stdio.h"int itoc(int num, char *buffer){ int i=0;if(num<0){ buffer='-';num=0-num;itoc(num,buffer+1);} else if(num>=10){ i=itoc(num/10,buffer);(buffer+i)=(num%10)+'0';(buffer+i+1)=0;} else ...