Trouble using fork() to calculate a total sum of received command line arguments.

Answer 1

Try this, if you want to change the pointers.

void ChangePointers(int **input, int **output)

and

ChangePointers(&input, &output);

Answer 2

Ray, it's hard to know what's wrong without seeing more details about the errors. If, as I suspect, these are run-time errors, can you run the code under valgrind? valgrind is extremely effective at pinpointing memory errors; with your application you'll want

valgrind --trace-children=yes ./coordinator 1 2 3 4

EDIT: OK, with the valgrind errors we can see that (a) you're passing something dodgy to printf (if you compile with -g you'll get the exact line number), and also you're calling realloc but not on a pointer you got back from malloc. Maybe you've done some pointer arithmetic?

Can't say more without seeing the code, but I hope you find valgrind helpful.

Answer 3

@Norman

This is the output I receive:

==3585== Memcheck, a memory error detector
==3585== Copyright (C) 2002-2009, and GNU GPL'd, by Julian Seward et al.
==3585== Using Valgrind-3.5.0 and LibVEX; rerun with -h for copyright info
==3585== Command: ./coordinator 1 2 3 4
==3585== 
calc: 2:
input[0]: 1
input[1]: 2
input[2]: 3
input[3]: 4
==3585== Use of uninitialised value of size 4
==3585==    at 0x4076186: ??? (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x4079A81: vfprintf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x4080F7F: printf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x8048833: main (in /home/bryan/cpp/coordinator)
==3585== 
==3585== Conditional jump or move depends on uninitialised value(s)
==3585==    at 0x407618E: ??? (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x4079A81: vfprintf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x4080F7F: printf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x8048833: main (in /home/bryan/cpp/coordinator)
==3585== 
==3585== Conditional jump or move depends on uninitialised value(s)
==3585==    at 0x4077877: vfprintf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x4080F7F: printf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x8048833: main (in /home/bryan/cpp/coordinator)
==3585== 
==3585== Conditional jump or move depends on uninitialised value(s)
==3585==    at 0x407789B: vfprintf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x4080F7F: printf (in /lib/tls/i686/cmov/libc-2.10.1.so)
==3585==    by 0x8048833: main (in /home/bryan/cpp/coordinator)
==3585== 
input[4]: 0
==3586== Memcheck, a memory error detector
==3586== Copyright (C) 2002-2009, and GNU GPL'd, by Julian Seward et al.
==3586== Using Valgrind-3.5.0 and LibVEX; rerun with -h for copyright info
==3586== Command: ./worker 1 2
==3586== 
Process ID: 3586 
Sum of 1 and 2 is 3 

==3586== 
==3586== HEAP SUMMARY:
==3586==     in use at exit: 0 bytes in 0 blocks
==3586==   total heap usage: 0 allocs, 0 frees, 0 bytes allocated
==3586== 
==3586== All heap blocks were freed -- no leaks are possible
==3586== 
==3586== For counts of detected and suppressed errors, rerun with: -v
==3586== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 11 from 6)
==3587== Memcheck, a memory error detector
==3587== Copyright (C) 2002-2009, and GNU GPL'd, by Julian Seward et al.
==3587== Using Valgrind-3.5.0 and LibVEX; rerun with -h for copyright info
==3587== Command: ./worker 3 4
==3587== 
Process ID: 3587 
Sum of 3 and 4 is 7 

==3587== 
==3587== HEAP SUMMARY:
==3587==     in use at exit: 0 bytes in 0 blocks
==3587==   total heap usage: 0 allocs, 0 frees, 0 bytes allocated
==3587== 
==3587== All heap blocks were freed -- no leaks are possible
==3587== 
==3587== For counts of detected and suppressed errors, rerun with: -v
==3587== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 11 from 6)
==3585== Invalid write of size 4
==3585==    at 0x8048A3A: main (in /home/bryan/cpp/coordinator)
==3585==  Address 0x417f0b4 is 8 bytes after a block of size 4 alloc'd
==3585==    at 0x4024C6C: malloc (vg_replace_malloc.c:195)
==3585==    by 0x4024CF6: realloc (vg_replace_malloc.c:476)
==3585==    by 0x8048A25: main (in /home/bryan/cpp/coordinator)
==3585== 
==3588== Memcheck, a memory error detector
==3588== Copyright (C) 2002-2009, and GNU GPL'd, by Julian Seward et al.
==3588== Using Valgrind-3.5.0 and LibVEX; rerun with -h for copyright info
==3588== Command: ./worker 3 7
==3588== 
Process ID: 3588 
Sum of 3 and 7 is 10 

==3588== 
==3588== HEAP SUMMARY:
==3588==     in use at exit: 0 bytes in 0 blocks
==3588==   total heap usage: 0 allocs, 0 frees, 0 bytes allocated
==3588== 
==3588== All heap blocks were freed -- no leaks are possible
==3588== 
==3588== For counts of detected and suppressed errors, rerun with: -v
==3588== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 11 from 6)
==3585== Invalid read of size 4
==3585==    at 0x8048AB5: main (in /home/bryan/cpp/coordinator)
==3585==  Address 0x417f0e0 is 0 bytes after a block of size 0 alloc'd
==3585==    at 0x4024C6C: malloc (vg_replace_malloc.c:195)
==3585==    by 0x4024CF6: realloc (vg_replace_malloc.c:476)
==3585==    by 0x8048A77: main (in /home/bryan/cpp/coordinator)
==3585== 
The final sum is: 0==3585== 
==3585== HEAP SUMMARY:
==3585==     in use at exit: 28 bytes in 2 blocks
==3585==   total heap usage: 4 allocs, 2 frees, 32 bytes allocated
==3585== 
==3585== LEAK SUMMARY:
==3585==    definitely lost: 8 bytes in 1 blocks
==3585==    indirectly lost: 0 bytes in 0 blocks
==3585==      possibly lost: 20 bytes in 1 blocks
==3585==    still reachable: 0 bytes in 0 blocks
==3585==         suppressed: 0 bytes in 0 blocks
==3585== Rerun with --leak-check=full to see details of leaked memory
==3585== 
==3585== For counts of detected and suppressed errors, rerun with: -v
==3585== Use --track-origins=yes to see where uninitialised values come from
==3585== ERROR SUMMARY: 6 errors from 6 contexts (suppressed: 11 from 6)

Answer 4

You have a couple of off-by-one errors:

1. for(i = 0; i < argc; i++)
2. while(calc > 0)

Answer 5

You should think about writing a function, let's call it step_once(), which will take input with n numbers, and write to the corresponding output with m = n/2 elements. n above is the number of input numbers + 1, with the last element of input equal to 0.

In your driver function, let's say main(): if output contains 1 number, you are done. Otherwise, you reallocate input to contain n_new = m+1 elements, reallocate output to contain m_new = n_new/2 elements, and call the function step_once() again. You keep doing this until you get one number:

function next_step(input, output, n, m):
    n := number of input numbers # this is 1 greater than
                                 # the number of numbers being summed
    m := n / 2 # C division
    n_children := m
    i := 0
    while i < m:
        fork worker with input[2*i] and input[2*i+1]
        get result in output[i]
        i := i + 1

function main:
    set n := length(input) + 1
    set m := n/2
    allocate memory for input # n+1 elements, last = 0
    allocate memory for output # m elements
    set values in input
    while True:
        next_step(input, output, n, m)
        if length or output == 1:
             done, return
        else:
            set n := length(output) + 1
            set m := n/2
            allocate space for new_input # n elements
            set new_input := output + [0]
            free input and output
            set input := new_input
            allocate memory for output # m elements

The advantage is that you can test your next_step() function to make sure it works and thus makes debugging easier.

An example will make this clearer:

Let's say you want to add 7 numbers: 1 2 3 4 5 6 7

1. input = [1 2 3 4 5 6 7 0]
2. n = len(input) = 8
3. m = n/2 = 4
4. output = [0 0 0 0]
5. We fork 4 process, first process gets [1 2], second gets [3 4], ...
6. The 4 processes return 3, 7, 11, 7 respectively, which we assign to output.
7. output has 4 elements, so we allocate space for 4+1 = 5 elements for the new input.
8. set input = [3 7 11 7 0]
9. n = len(input) = 5
10. m = n/2 = 2
11. output = [0 0]
12. We fork 2 processes, first gets [3 7], second gets [11 7]
13. The 2 processes return 10, 18, which we assign to output.
14. output has 2 elements, so we allocate space for 2+1 = 3 elements for the new input.
15. set input = [10 18 0]
16. n = len(input) = 3
17. m = n/2 = 1
18. output = [0]
19. We fork one process, which gets [10 18]
20. The process returns 28, which we assign to output.
21. output has 1 element, so we are done.