I can’t divide with a vector element

Question

I made a code to simulate the Monty Hall paradox, so that I can simulate 1000 times the results and write the statistics. However, I cannot divide an element of an int array by an int, every time I run the program, when it reaches the division, the program crashes and has to be closed:

void main(){
    int PORTAS [] = {0, 0, 0}; 
    int CAR, ESC, NUM, DECISAO, AUX;
    float VIT, DER; 
    int ESTAT [] = {0, 0, 0};   // {Tentativas, vitórias, derrotas}

    VIT = ( ESTAT[2] / NUM ) * 100; // Porcentagem de vitórias  *ERRO*
    DER = ( ESTAT[3] / NUM ) * 100; // Porcentagem de derrotas  *ERRO*
}

I also tried to write how:

VIT = ESTAT[2]/ESTAT[1]*100;
DER = ESTAT[3]/ESTAT[1]*100;

But in this way, the result is always 0.

If necessary to help me, I will put the full code, which includes at the beginning a brief explanation of the game. If you don’t think it’s necessary, ignore the code below:

#include <stdio.h>
#include <stdlib.h>
#include <locale.h>
#include <math.h>

/*Paradoxo de Monty Hall
*Atrás de uma porta, de três no total, é escondido um carro
*e nas outras duas, uma cabra. No jogo, o jogador escolhe uma
*porta e então é revelada outra porta que contenha uma cabra.
*O jogador tem então a opção de escolher entre manter a escolha
*ou trocar de porta. Esse programa simula esse jogo, de modo que 
*a porta seja mantida ou trocada em todas as tentativas, para
*motivos estatísticos.*/

void main(){
    setlocale (LC_ALL, "");

    int PORTAS [] = {0, 0, 0};  // 0 - a porta contém uma cabra; 1 - a porta contém o carro 
    int CAR, ESC, NUM, DECISAO, AUX;
    float VIT, DER; 
    int ESTAT [] = {0, 0, 0};   // {Tentativas, vitórias, derrotas}

    printf("1 para trocar todas, 0 para manter todas: ");
    scanf ("%d", &DECISAO);
    printf ("Digite o número de repetições: "); // Recomendável 10, 100 ou 1000.
    scanf ("%d", &NUM);

    do{
        CAR = rand () %3;   // Randomiza a porta que recebe o carro.
        PORTAS [CAR] = 1;   

        ESC = rand () %3;   // Randomiza a escolha da porta.

        if ( DECISAO == 1 ){    // Se foi escolhido trocar todas as vezes.
            if ( PORTAS [ESC] == 1 ){   // Porta escolhida contém o carro.
                for ( AUX = 0; AUX < 3 ; AUX++ ){
                    if ( PORTAS [AUX] != 1 && AUX != ESC ){
                        ESC = AUX;      //Mudança de porta
                        AUX = 3;    // Para quebrar o 'For'
                        ESTAT [3] += 1;
                    }
                }
            }

            if ( PORTAS [ESC] == 0){    //Porta escolhida contém uma cabra.
                ESC = CAR;  // Pois sendo a porta errada, e tendo a outra errada sido revelada, só sobrou a correta.
                ESTAT [2] += 1;
            }
        }

        if ( DECISAO == 0){     //Caso tenha sido escolhido manter todas as vezes.
            if ( ESC = CAR ){
                ESTAT [2] += 1;
            }
            else{
                ESTAT [3] += 1;
            }
        }

        NUM--;
        ESTAT [1] += 1;
    } while ( NUM > 0);

    VIT = ( ESTAT[2] / NUM ) * 100; // Porcentagem de vitórias  *ERRO*
    DER = ( ESTAT[3] / NUM ) * 100; // Porcentagem de derrotas  *ERRO*

    ( DECISAO == 1 ) ? printf ("\n\n\n\tTrocando de porta todas as vezes: \n\n") : printf ("\n\n\n\tMantendo a porta todas as vezes: \n\n");
    printf ("Número de tentativas: %d\n", ESTAT [1]);
    printf ("Número de vitórias: %d, %d%% do total.\n", ESTAT [2], VIT);
    printf ("Número de derrotas: %d, %d%% do total.", ESTAT [3], DER);

    getch();
}

Answer 1

I don’t know what the goal is but this is basic math. The code is running when NUM is 0 and this operation is not possible according to the mathematical rule.

Probably shouldn’t wear one array For this specific question problem, three variables would be more readable. This is just one of the things that makes code difficult to read and understand. In addition it is using elements 1, 2 and 3 when it should be 0, 1 and 2.

The second way may not cause the problem but is it what you want? You can’t keep including random variables in the expression to see if it works. And this used is very random, it doesn’t make any sense. It has to do what it needs. And only you know this.

I think it should be like in the first form, but keeping the initial value of the variable. It cannot reuse variables. I created a new variable called repeticoes to store the value and make the division correctly.

Actually the code is full of errors. It is no use trying to solve complex problems when you do not know the basics of programming. After solving several compilation errors it seems that worked, but I do not guarantee that you are doing what you want. So there must be problems that go beyond what is in the question.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main() {
    int PORTAS[] = {0, 0, 0};  // 0 - a porta contém uma cabra; 1 - a porta contém o carro 
    int ESTAT[] = {0, 0, 0};   // {Tentativas, vitórias, derrotas}
    int repeticoes;
    int DECISAO;
    printf("1 para trocar todas, 0 para manter todas: ");
    scanf ("%d", &DECISAO);
    printf ("Digite o número de repetições: "); // Recomendável 10, 100 ou 1000.
    scanf ("%d", &repeticoes);
    int NUM = repeticoes;
    do {
        int AUX;
        int CAR = rand () %3;   // Randomiza a porta que recebe o carro.
        PORTAS[CAR] = 1;   
        int ESC = rand () %3;   // Randomiza a escolha da porta.
        if (DECISAO == 1) {    // Se foi escolhido trocar todas as vezes.
            if (PORTAS[ESC] == 1 ) {   // Porta escolhida contém o carro.
                for (AUX = 0; AUX < 3 ; AUX++) {
                    if (PORTAS[AUX] != 1 && AUX != ESC) {
                        ESC = AUX;      //Mudança de porta
                        AUX = 3;    // Para quebrar o 'For'
                        ESTAT[2] += 1;
                    }
                }
            }
            if (PORTAS[ESC] == 0) {    //Porta escolhida contém uma cabra.
                ESC = CAR;  // Pois sendo a porta errada, e tendo a outra errada sido revelada, só sobrou a correta.
                ESTAT[1] += 1;
            }
        }
        if (DECISAO == 0) {     //Caso tenha sido escolhido manter todas as vezes.
            if (ESC == CAR ) ESTAT[1] += 1;
            else  ESTAT[2] += 1;
        }
        NUM--;
        ESTAT[0] += 1;
    } while (NUM > 0);
    float VIT = (ESTAT[2] / repeticoes) * 100; // Porcentagem de vitórias  *ERRO*
    float DER = (ESTAT[3] / repeticoes) * 100; // Porcentagem de derrotas  *ERRO*
    (DECISAO == 1) ? printf("\n\n\n\tTrocando de porta todas as vezes: \n\n") : printf("\n\n\n\tMantendo a porta todas as vezes: \n\n");
    printf("Número de tentativas: %d\n", ESTAT[0]);
    printf("Número de vitórias: %d, %f%% do total.\n", ESTAT[1], VIT);
    printf("Número de derrotas: %d, %f%% do total.", ESTAT[2], DER);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

I improved the code but not everything.

Answer 2

I also tried to write how:

VIT = ESTAT[2]/ESTAT[1]*100;
DER = ESTAT[3]/ESTAT[1]*100;

There are two problems here: first, in C the arrays are indexed from zero, not from one, so that the first element has index 0, the second index 1 and the third index 2. Access the index 3 how are you doing is invalid for a array size 3, and the result is undefined (because you are accessing an arbitrary memory region, which may even be outside the scope of your program).

The second problem is that the entire division will always give an entire result, even if the variable where you intend to store the result is float. If the dividend is less than the divisor then the result will always be zero (as you observed). Fixing both problems we have:

 VIT = 1.0f*ESTAT[1]/ESTAT[0]*100; // Ao multiplicar primeiro por 1.0f
 DER = 1.0f*ESTAT[2]/ESTAT[0]*100; // o operando passa a ser um float

Already his first attempt (as NUM) was flawed because you’ve been decreasing NUM in his loop down to zero. When trying to divide by zero, naturally there would be a problem...

P.S. I didn’t revise the full code - because Maniero has already done it - but after the correction of the main errors we arrived to that final result (the result of which corresponds to the expected, statistically speaking). Some recommendations:

• Do not use full-name variables in upper case, as this notation is traditionally used to represent constants only;
• Document your variables, and try to give them self-descriptive names. I personally had no trouble understanding what it was CAR[ro] and ESC[look], but this is due to the previous knowledge I had from Monty Hall, other people may have more difficulty relating the variables to the elements of the domain;

• If two conditions are mutually exclusive, I suggest using else even if you don’t think it’s necessary. In this code:

      if ( PORTAS [ESC] == 1 ){   // Porta escolhida contém o carro.
         ...
      }
  
      if ( PORTAS [ESC] == 0){    //Porta escolhida contém uma cabra.
         ...
      }
  

  you didn’t use the else, and so he ended up entering the second if even after having entered the first (because this modified ESC) - causing the number of hits plus the number of errors to exceed the number of attempts...

  Two ifs in a row without else should only be used if there is a chance of entering: a) only in the first; b) only in the second; c) in both; d) in either.

• I saw that you tried to simulate the change of choice (reassigning ESC), but in this case this is unnecessary - if he chose right, he will lose, if he chose wrong, he will win, there is no need to simulate anything else. Anyway, your loop who uses AUX is correct, although a bit confused (by the way, you knew you could leave early from a loop using break?).