Question

This program should calculate the value of 2^1+2^2 + … + 2^10:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <stdbool.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <signal.h>
#include <math.h>

#define N 10

// sommatoria per i che va da 1 a N di 2^i, ogni processo calcola un singolo valore

int main(int argc, char** argv)
{
    pid_t figli[N];
    unsigned int i;
    int status;
    int fd[N][2];
    int msg1=0,msg2;
    int risultato=0;
    bool padre=true;
    for(i=0;i<N && padre;i++)
    {
        pipe(fd[i]);
        figli[i]=fork();
        if(figli[i]<0)
        {
            fprintf(stderr,"Una fork ha fallito\n");
        }
        else if(figli[i]==0)
        {
            padre=false;
        }
        else
        {
            msg1=i+1;
            write(fd[i][1],&msg1,sizeof(int));
        }
    }
    if(!padre)
    {
        read(fd[i][0],&msg2,sizeof(int));
        msg2=pow(2.0,msg2);
        write(fd[i][1],&msg2,sizeof(int));
        exit(0);
    }
    else
    {
        for(i=0;i<N;i++)
        {
            read(fd[i][0],&msg2,sizeof(int));
            risultato+=msg2;
        }
    }
    if(padre)
        fprintf(stderr,"%d\n",risultato);
    return 0;
}

But when ie xecute the program, the father process prints 55.
Why?

Answer 1

Interestingly enough, 55 is the sum of all the numbers from 1 to 10: that should give you an instant clue:

pipe() creates a pipe, a unidirectional data channel that can be used for interprocess communication. The array pipefd is used to return two file descriptors referring to the ends of the pipe. pipefd[0] refers to the read end of the pipe. pipefd[1] refers to the write end of the pipe.

Note that well: unidirectional. In other words, the padre is reading back the same values it wrote (hence the 55).

You normally set up two pipes for bi-directional traffic, one for each direction. So I’ve double the number of pipes, using even ones for padre-to-child and odd ones for the other direction.

In addition, your children continue with the padre loop whereas they should exit that loop immediately so their value of i is correct. You do have the loop exit based on padre but this happens after i has changed. You can either break where you set padre to false or simply i-- in the if(!padre) bit to restore i to the correct value for this child. I’ve done the latter.

The following code (with markers showing what changed) works okay:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <stdbool.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <signal.h>
#include <math.h>

#define N 10

int main(int argc, char** argv)
{
    pid_t figli[N];
    unsigned int i;
    int status;
    int fd[N*2][2];  // CHANGED: two unidirectional pipes
    int msg1=0,msg2;
    int risultato=0;
    bool padre=true;
    for(i=0;i<N && padre;i++)
    {
        pipe(fd[i*2]);
        pipe(fd[i*2+1]); // ADDED: create second pipe
        figli[i]=fork();
        if(figli[i]<0)
        {
            fprintf(stderr,"Una fork ha fallito\n");
        }
        else if(figli[i]==0)
        {
            padre=false;
        }
        else
        {
            msg1=i+1;
            write(fd[i*2][1],&msg1,sizeof(int));  // CHANGED: pipe number
        }
    }
    if(!padre)
    {
        i--;  // ADDED: to restore i for the child
        read(fd[i*2][0],&msg2,sizeof(int));  // CHANGED: pipe number
        msg2=pow(2.0,msg2);
        write(fd[i*2+1][1],&msg2,sizeof(int));  // CHANGED: pipe number
        exit(0);
    }
    else
    {
        for(i=0;i<N;i++)
        {
            read(fd[i*2+1][0],&msg2,sizeof(int));  // CHANGED: pipe number
            risultato+=msg2;
        }
    }
    if(padre)
        fprintf(stderr,"%d\n",risultato);
    return 0;
}

This generates the correct answer of 2046, since 20 + 21 + ... 210 = 211 - 1 and, since you’re leaving out the two-to-the-zero term (equal to 1): 21 + 22 + ... 210 is 211 - 2 (211 = 2048).