在递归函数中使用 fork

Question

I have a recursive function, in which I make a pipe. I am assuming that each function call has it's own pipe. Is that the case?

Here is my recursive function.

int sum(tree *&root , tree *&root2, int value1 , int value2 , int result_tree)
{
    if(root->l_child == NULL && root->r_child == NULL)
        return root->data;
    else
    {
        value1 = sum(root->l_child , root2 , value1 , value2 , result_tree);
        value2 = sum(root->r_child , root2 , value1 , value2 , result_tree);

        int fd2[2];
        pipe(fd2);
        if(fork() == 0)
        {
            if(root->data == 43)
                result_tree = value1 + value2;      
            else if(root->data == 45)
                result_tree = value1 - value2;  
            else if(root->data == 42)
                result_tree = value1 * value2;  
            else if(root->data == 47)
                result_tree = value1 / value2;  

            close(fd2[0]);
            write(fd2[1] , &result_tree , sizeof(result_tree));
        }
        else
        {
            close(fd2[1]);
            read(fd2[0] , &result_tree , sizeof(result_tree));
        }

        root->data = result_tree;
        delete_node(root2 , root);
        //cout<<"\n";
        //display_in(root2);
        return result_tree;
    }
}

It gives output as much times as is the number of operands in the expression.For example, if the user enters expression: 3+4+4 then the output is: result:11result11result11

what is wrong init.

Answer 1

In response to your question

I am assuming that each function call has it's own pipe. Is that the case?

You are calling the function pipe once per invocation of your function sum . This creates a unique,anonymous,bidirectional pair of pipes. However, you are not checking the return value of pipe . It is possible the pipes are not being created because you are ignoring the return value. The man page states

On success, zero is returned.

You should check that pipe returns zero when called, and abort if zero is not returned. The same goes for calls to close , write , and read .

If you include in your question your goal and why you think your code does not accomplish it, it is possible to provide a more informed answer. Just looking at your code, I am assuming that you understand that when the result of fork is zero, you are in the child process. It looks like you are writing some results to a pipe, but then you continue executing. This probably is not your intent. If it was not, you should call exit in the child process to terminate execution after writing to the pipe.

Answer 2

got the code fixed! I just included exit(0) in child process and it's working well.

int sum(tree *&root , tree *&root2, int value1 , int value2 , int result_tree)
{
     if(root->l_child == NULL && root->r_child == NULL)
         return root->data;
     else
     {
         value1 = sum(root->l_child , root2 , value1 , value2 , result_tree);
         value2 = sum(root->r_child , root2 , value1 , value2 , result_tree);

         int fd2[2];
         pipe(fd2);
         if(fork() == 0)
         {
             if(root->data == 43)
                 result_tree = value1 + value2;      
             else if(root->data == 45)
                 result_tree = value1 - value2;  
             else if(root->data == 42)
                 result_tree = value1 * value2;  
             else if(root->data == 47)
                 result_tree = value1 / value2;  

             close(fd2[0]);
             write(fd2[1] , &result_tree , sizeof(result_tree));
             exit(0);                                                       // change
         }
         else
         {
             close(fd2[1]);
             read(fd2[0] , &result_tree , sizeof(result_tree));
         }

         root->data = result_tree;
         delete_node(root2 , root);
         //cout<<"\n";
         //display_in(root2);
         return result_tree;
     }
}

I don't understand this that what actually exit(0) did , so that the code gave correct output. Can anyone explain this exit(0) thing , in this particular piece of code. Thankyou!