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Clean ways to do multiple undos in C

Tags:

c

Someone will probably say something about exceptions... but in C, what are other ways to do the following cleanly/clearly and without repeating so much code?

if (Do1()) { printf("Failed 1"); return 1; }
if (Do2()) { Undo1(); printf("Failed 2"); return 2; }
if (Do3()) { Undo2(); Undo1(); printf("Failed 3"); return 3; }
if (Do4()) { Undo3(); Undo2(); Undo1(); printf("Failed 4"); return 4; }
if (Do5()) { Undo4(); Undo3(); Undo2(); Undo1(); printf("Failed 5"); return 5; }
Etc...

This might be one case for using gotos. Or maybe multiple inner functions...

like image 661
dargaud Avatar asked Nov 23 '18 10:11

dargaud


3 Answers

Yes, it's quite common to use goto in such cases to avoid repeating yourself.

An example:

int hello() {
  int result;

  if (Do1()) { result = 1; goto err_one; }
  if (Do2()) { result = 2; goto err_two; }
  if (Do3()) { result = 3; goto err_three; }
  if (Do4()) { result = 4; goto err_four; }
  if (Do5()) { result = 5; goto err_five; }

  // Assuming you'd like to return 0 on success.
  return 0;

err_five:
  Undo4();
err_four:
  Undo3();
err_three:
  Undo2();
err_two:
  Undo1();
err_one:
  printf("Failed %i", result); 
  return result;
}

As always you probably also want to keep your functions small and batch together the operations in a meaningful manner to avoid a large "undo-code".

like image 93
likle Avatar answered Oct 23 '22 04:10

likle


This might be one case for using gotos.

Sure, let's try that. Here's a possible implementation:

#include "stdio.h"
int main(int argc, char **argv) {
    int errorCode = 0;
    if (Do1()) { errorCode = 1; goto undo_0; }
    if (Do2()) { errorCode = 2; goto undo_1; }
    if (Do3()) { errorCode = 3; goto undo_2; }
    if (Do4()) { errorCode = 4; goto undo_3; }
    if (Do5()) { errorCode = 5; goto undo_4; }

undo_5: Undo5();    /* deliberate fallthrough */
undo_4: Undo4();
undo_3: Undo3();
undo_2: Undo2();
undo_1: Undo1();
undo_0: /* nothing to undo in this case */

    if (errorCode != 0) {
        printf("Failed %d\n", errorCode);
    }
    return errorCode;
}
like image 18
Blaze Avatar answered Oct 23 '22 04:10

Blaze


If you have the same signature for your function you can do something like this:

bool Do1(void) { printf("function %s\n", __func__); return true;}
bool Do2(void) { printf("function %s\n", __func__); return true;}
bool Do3(void) { printf("function %s\n", __func__); return false;}
bool Do4(void) { printf("function %s\n", __func__); return true;}
bool Do5(void) { printf("function %s\n", __func__); return true;}

void Undo1(void) { printf("function %s\n", __func__);}
void Undo2(void) { printf("function %s\n", __func__);}
void Undo3(void) { printf("function %s\n", __func__);}
void Undo4(void) { printf("function %s\n", __func__);}
void Undo5(void) { printf("function %s\n", __func__);}


typedef struct action {
    bool (*Do)(void);
    void (*Undo)(void);
} action_s;


int main(void)
{
    action_s actions[] = {{Do1, Undo1},
                          {Do2, Undo2},
                          {Do3, Undo3},
                          {Do4, Undo4},
                          {Do5, Undo5},
                          {NULL, NULL}};

    for (size_t i = 0; actions[i].Do; ++i) {
        if (!actions[i].Do()) {
            printf("Failed %zu.\n", i + 1);
            for (int j = i - 1; j >= 0; --j) {
                actions[j].Undo();
            }
            return (i);
        }
    }

    return (0);
}

You can change the return of one of Do functions to see how it react :)

like image 14
Tom's Avatar answered Oct 23 '22 04:10

Tom's