marlonbymendes's blog

Using memset on array of bool (C++)
By marlonbymendes, history, 8 years ago, In English

The following piece of code:

    const int N = 10;
    bool can[N];
    memset(can, -1, sizeof can);

    if(can[0] == 0) { // can[0] == false gives same result
        cout << "is zero";
    }
    else {
        cout << "not zero";
    }

Outputs is zero, however I expected it to be not zero.

Using memset like this instead:

    memset(can, 1, sizeof can); // using true instead of 1 gives same result

Outputs not zero as expected. What is going on?

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8 years ago, # |
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for me, first one prints not zero as expected. I checked it using codeforces custom test as well, again not zero.

basically memset changes every 8 bits of the memory, to the value, so for types like char which are 8-bits, it simply fills the array with the value. 

char a[10];
memset(a, 'a', sizeof a);

now all of a is 'a'

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    8 years ago, # ^ |
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    It seems that sizeof(char) is defined to be 1. sizeof(bool) however is said to be implementation-defined, wich may explain we have different results. A friend of mine tested the above code, having the same result (is zero) as me and I still have no explanation for what is going on internally.

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      8 years ago, # ^ |
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      The problem with your code is if (can[0] == 0). If you change it to if (!can[0]) I'm pretty sure that it would run as supposed to run.

      Edit:

      and it doesn't matter what sizeof bool is, when you memset it with -1, it's gonna be all set bits cause -1 is 1111111.

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        8 years ago, # ^ |
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        You're right, i just test it and it worked as I expected using if(!can[0]) and if(can[0]).

        However, the code below prints both not zero and is zero, which blows my mind and I have no idea what is going on internally because bool is defined as 0, which should compare true to == 0.

        const int N = 10;
    bool can[N];
    memset(can, -1, sizeof can);

    if(can[0] != 0) {
        cout << "not zero";
    }
    else {
        cout << "is zero";
    }
    cout << "\n";

    if(can[0] == 0) {
        cout << "is zero";
    }
    else {
        cout << "not zero";
    }
    cout << "\n";

        As programmers we expect that something which compares true to (a != 0) should compare false to (a == 0).

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8 years ago, # |
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What's the compiler you've used?

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    8 years ago, # ^ |
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    From g++ --version: "g++ (Ubuntu 5.4.0-6ubuntu1~16.04.4) 5.4.0 20160609"

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      8 years ago, # ^ |
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      g++ assumes the value of bool is 0 or 1,if you look at the assemly you will find it is actually testing can[0]^1. Here, can[0] is 255 and 255^1=254, which is evaluated as true.

      The above conclusion is based on default config of g++ on several Linux platforms(simplely g++ a.cpp -o a), result may change under different environment.

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        8 years ago, # ^ |
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        I'm sorry but how can 254 == 0 compare true? Or if it's either 1 or 0, how can it compare both true to != 0 and false to == 0? Please read my answer above to HosseinYousefi for details.

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          8 years ago, # ^ |
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          Not 254==0, 254 is true.

          if(can[0]==0)->if(can[0]^1)->if(254)

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8 years ago, # |
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You may want to read: Representation of Integers and Reals

For this particular problem, read the section headed with Rumor: In C++, the code “int A[1000]; memset(A,x,sizeof(A));” stores 1000 copies of x into A.

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