Doubt in vector in c

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-1

Create a counter function that works as follows:

  • the initial value of the counter is 5.
  • every time the function is called its counter shall be decreased.
  • the counter shall reset automatically: when the counter reaches 0, on the next function call the counter shall be reset with 5 again.
#include<stdio.h>
static int contador = 5;

void cont();

int main(){
        cont();
        return 0;
}

void cont(){
        int i,j=0;
        int vet[contador];
        do{
                for(i = 1;i <= contador;i++){
                        vet[i]++;
                        printf("%d\n",vet[i]);
                }
                contador -=1;
        }while(contador != 0);

}
c logic
  • 1

    Eae man, you could explain your doubt?

  • I am doubtful on the part of implementing an array and when the counter should reset automatically when the counter reaches 0.

  • 1

    Your question says nothing about vector, where a vector enters what you want?

  • then and bury the vector and consider that my doubt and the counter should reset automatically: when the counter reaches 0, on the next call of the function the counter should be reset with 5 again.

2 answers

3


There is a problem in its function, it accesses an invalid memory position. In C the indexes of a vector go from 0 to N-1. In your case the vector goes from v[0] to v[4], need to fix this in the loop.

To make the counter restart to 5 just put one if at the end of the loop. Note that with this your program will go into infinite loop, since counter will always be different from 0 when it is evaluated on do{}while(). Follows the amendment below:

void cont(){
        int i;
        int vet[contador];
        do{
                for(i = 0; i < contador; i++)
                {
                        vet[i]++;
                        printf("%d\n",vet[i]);
                }
                contador -=1;
                if (contador == 0)
                    contador = 5;
        }while(contador != 0);
}
  • 2

    The if should go outside the do-while, otherwise the loop will be infinite, because whenever it reaches the end, the value of contador will again be 5 and the condition contador != 0 will never be satisfied.

1

From what I can understand, you want to make a countdown count of how many times this function was called, arriving at the value 0 (zero) it returns to the initial value.

#include <stdio.h>

#define MAX 3

int GLOBAL_COUNT = MAX;

void contador();

int main() {
    for(int i = 0; i <= 7; i++) {
        contador();
    }
    return 0;
}

void contador() {
    if (GLOBAL_COUNT <= 0) {
        GLOBAL_COUNT = MAX;
    }
    printf("chamada: %d/%d\n", GLOBAL_COUNT, MAX);
    GLOBAL_COUNT--;
}

See the example working on IDEONE.

In this my example I took the liberty of changing the value to 3 just to simplify the visualization and just that, in it I created a constant value MAX and assigns to a variable that is the global replay counter, this will be decremented in the function call.

In the main I made a loop of 7 calls to show the quoter working.

In accountant() it is comparing whether the current value of GLOBAL_COUNT is less than or equal to 0 (zero), if so, then it restarts the value, after which it will print the current value and then it will decrement stating that it passed the function.

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