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Joined 1 year ago
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Cake day: July 3rd, 2023

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  • It doesn’t violate any rules… Imagine both the “speaker” and the “text” are being updated by separate threads. A program that would eventually display the behavior in this meme is simple, and I’m a bit embarrassed to have written it because of this comment:

    #include <pthread.h>
    #include <stdio.h>
    
    char* speakers[] = {
        "Alice",
        "Bob"
    };
    int speaker = 0;
    
    void* change_speaker(void* arg)
    {
        (void)arg;
    
        for (;;) {
            speaker = speaker == 0 ? 1 : 0;
        }
    }
    
    char* texts[] = {
        "Hi Bob",
        "Hi Alice, what's up?",
        "Not much Bob",
    };
    int text = 0;
    
    void* change_text(void* arg)
    {
        (void)arg;
        for (;;) {
            switch (text) {
            case 0:
                text = 1;
                break;
            case 1:
                text = 2;
                break;
            case 2:
                text = 0;
                break;
            }
        }
    }
    
    int main(int argc, char* argv[])
    {
        pthread_t speaker_swapper, text_swapper;
    
        pthread_create(&text_swapper, NULL, change_text, NULL);
        pthread_create(&speaker_swapper, NULL, change_speaker, NULL);
        for (int i = 0; i < 3; ++i) {
            printf("%s: %s\n", speakers[speaker], texts[text]);
        }
    }
    




  • This is not necessarily true.

    For example, consider the case of a 1Password vault falling into the hands of an attacker. They do not have the option to just crack your password, as the password is mixed with a randomly generated value to ultimately derive the key. They would need to simultaneously brute force your password and that random value. This should almost be impossible. However, given access to a client that already has knowledge of the secret value, it would fall back to brute forcing the password.