C语言 用递归函数求数值的整数次幂 double power（double x,int p）输入负整数次幂时出现问题

C语言 用递归函数求数值的整数次幂 double power（double x,int p）输入负整数次幂时出现问题代码如下：#includedouble power_positive(double n,int p);double power_negative(double n,int p);int main(void){double x,xpow;int exp;printf("Enter a number and the integer power to which\n");printf("the number will be raised.Enter q to quit.\n");while(scanf("%lf %d",&x,&exp)==2){if(x==0)printf("%lf to the power of %d is 0.\n",x,exp);else if(exp==0)printf("%lf to the power of %d is 1.\n",x,exp);else if(exp>0){xpow=power_positive(x,exp);printf("%lf to the power of %d is %lf.\n",x,exp,xpow);}else{xpow=power_negative(x,exp);printf("%lf to the power of %d is %lf.\n",x,exp,xpow);}printf("Enter next pair of numbers or q to quit.\n");}printf("Hope you enjoyed this power trip --BYE!\n");return 0;}double power_positive(double n,int p){double pow=1;if(p>0)pow=n*power_positive(n,(p-1));return pow;}double power_negative(double n,int p){double pow=1;double re;int q;q=-p;if(q>0)pow=n*power_negative(n,q-1);re=1/pow;return re;}可以运行.编译不报错.但是输入负整数次幂时计算结果根本就不对.也不知道错到哪里了.请高人指教哈.比如说2的-2次幂是0.52的-3次幂还是0.5一直都是0.5..