C学习02函数里面调用函数递归

#include <stdio.h>

int f(int x)
{
    printf("f--(%d)--%p\n",x,&x);
    if(x<=0){
         printf("f++(%d)++%p\n",x,&x);
        return 5;
    }else{
          int num=f(x-1);
          printf("return from f(%d) the num is:%d\n",x-1,num);
         printf("f**(%d)**%p\n",x,&x);
    // return 2*f(x-1)+3;//2*f(3-1)+3=2*f(2)+3 => 2*f(2-1)+3 => 2*f(1-1)+3=2*f(0)+3;属于递推           2*f(2)+3               2*f(1)+3      2*f(0)+3
        return 2*num+3;
    }
}
int main()
{
    printf("%d\n",f(3));
    return 0;
}
运算结果
f--(3)--0xbfb696f0
f--(2)--0xbfb696c0
f--(1)--0xbfb69690
f--(0)--0xbfb69660
f++(0)++0xbfb69660
return from f(0) the num is:5
f**(1)**0xbfb69690
return from f(1) the num is:13
f**(2)**0xbfb696c0
return from f(2) the num is:29
f**(3)**0xbfb696f0
61