c将指针传递给递归函数

Question

So here i am struggling with this program, i was trying to find out how can I use a array of pointers declared into main, in a recursive function to memorize data, the question that arises here is if it's the same approach as for a single pointer, what about for a struct type ? what is the best way to pass by reference a variable/array to a recursive function ?

#include <stdio.h>
#include <stdlib.h>
#define N 1

void f(int i,int j,int *cnt);

int j=0;

int main(int argc, char *argv[])
{
  int *cnt=0;
  f(0,++j,&cnt);
  printf("------ %d ---- \n",cnt);
  system("PAUSE");  
  return 0;
}

void f(int i,int j,int *cnt){

   if(i>N){
          printf("---if --- %d ---- %d \n",i,j);
          (*cnt)++;
          return;
          }

   (*cnt)++;
   printf("---bg --- %d ---- %d \n",i,j);
   f(i+1,++j,cnt);
   f(i+1,++j,cnt);        
}

Another thing i'd like to know is how does the recursive functions handle the ++i and i++ and i+1 increments (when passed as parameters),

Answer 1

int main(int argc, char *argv[])
{
    int *cnt=0;
    f(0,++j,&cnt);
    printf("------ %d ---- \n",(*cnt));
    system("PAUSE");  
    return 0;
}

needs to be

int main(int argc, char *argv[])
{
    int intStorage = 0;//<---- As Oli said.
    int *cnt= &intStorage;
    f(0,++j,cnt);//<-------AMPERSAND removed, overly dereferenced.
    printf("------ %d ---- \n",(*cnt));
    system("PAUSE");  
    return 0;
}

++i and i++ and i+1 (when passed as parameters):

1. ++i: i + 1 is passed and is also the value i takes afterwards.
2. i++: i is passed and i = i + 1 after the call.
3. i+1: i + 1 is passed but i remains as just i afterwards.

I'll try and fix your function a little too:

void f(int i,int j,int *cnt){

    if(i>N){
        printf("---if --- %d ---- %d \n",i,j);
        return;
    }

    (*cnt)++;
    printf("---bg --- %d ---- %d \n",i,j);
    if ( i < 50 && j < 50 ) {
        f(i+1,++j,cnt);
        f(i+1,++j,cnt);
    }
}

Still a lot of recursion but without the danger of not stopping.

Answer 2

A simple way of handling the pointer into your function is:

#include <stdio.h>
#include <stdlib.h>
#define N 1

void f(int i, int j, int *cnt);

int j = 0;

int main(void)
{
    int cnt = 0;
    f(0, ++j, &cnt);
    printf("------ %d ----\n", cnt);
    return 0;
}

void f(int i, int j, int *cnt)
{
   // Having a local variable j and a global j is likely to confuse someone!
   if (i > N)
   {
      printf("---if --- %d ---- %d\n", i, j);
      return;
   }

   (*cnt)++;
   printf("---bg --- %d ---- %d\n", i, j);
   f(i+1, ++j, cnt);
   f(i+1, ++j, cnt);
}

This code produces the following output with no crash:

---bg --- 0 ---- 1
---bg --- 1 ---- 2
---if --- 2 ---- 3
---if --- 2 ---- 4
---bg --- 1 ---- 3
---if --- 2 ---- 4
---if --- 2 ---- 5
------ 3 ----