Exercise with odd pairs in C

Question

Statement - Make an algorithm that reads a set of numbers (X) and indicate the amount of even numbers (Qpares) and the amount of odd numbers (Qimpares) read.

Note: Entering values ends when the user inserts 0.

      #include <stdio.h>
      main(){
      int num,res,qpar=0,qimp=0;
      while(num!=0){
      printf("Introduza");
      scanf("%d",&num);
        res=num%2;
       if(res==0){
        qpar++;
    } else {
        qimp++;
    }
   }
  printf("Pares -> %d\n",&qpar);
  printf("Impares -> %d\n",&qimp);
}

I don’t know what I’m doing wrong here, it compiles well but the problem is it doesn’t give me the right values of even and odd numbers. Otherwise it is also making the cycle correctly.

Answer 1

There are errors, the & is not used for printing. One more condition to validate before zero so that it does not enter as a pair.

Code

  #include <stdio.h>
  int main(){
      int num,qpar=0,qimp=0;
      float res;
      do{
          printf("Introduza");
          scanf("%d",&num);
          if(num!=0){ //Condicional para que  o 0 não conte como par    
              res=num%2;        

              if(res==0){
                  qpar++;
              } else {
                  qimp++;
              }
          }
      }while((num!=0));
      printf("Pares -> %d\n", qpar); //Retirado o &
      printf("Impares -> %d\n", qimp); //Retirado o &
  }

Answer 2

Your mistake is here:

    printf("Pares -> %d\n", &qpar);
    printf("Impares -> %d\n", &qimp);

Take these off &. You do not want to show the addresses of the variables, but the values. What you want is this:

    printf("Pares -> %d\n", qpar);
    printf("Impares -> %d\n", qimp);

Answer 3

Organizing the code helps to identify the problem. In the specific case it is having an integer pointer printed instead of an integer. If it were compiled as it should, it would not compile. and an error would already be presented.

The scanf() needs to pass a pointer, so uses the operator &(address of) because it needs to tell where the read value of the keyboard will be stored. no printf() should not do this, it is the very value you want to print and not the address where it is.

#include <stdio.h>

int main() {
    int num = 1, qpar = 0, qimp = 0;
    while (num != 0) {
        printf("Introduza\n");
        scanf("%d", &num);
        if (num % 2 == 0) {
            qpar++;
        } else {
            qimp++;
        }
    }
    printf("Pares -> %d\n", qpar);
    printf("Impares -> %d\n", qimp);
}

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

Answer 4

I can identify a new error in your code, you do not start the variable num before starting to check the while. This can be a problem because the memory region referenced by the variable can contain 0, if applicable, your while will be ignored.

I see two solutions:

• Swap your repetition structure for a do while.

• Simply perform a num scanf before entering the while. The printf error has already been identified, but remember that when you put & in front of a variable you are asking for its memory address.

I don’t know if you did it to post here, but it’s always good to separate some lines with white spaces, where it makes sense. Separating all the code can leave it as messy as not separating

Good luck, Keep coding.

Answer 5

#include <stdio.h>

void ex_3()
{
    int n,c=1;
    do
    {
        printf("DIGITE UM NUMERO INTEIRO: ");
        scanf("%d", &n);
        if(n<=0)
        {
            c=0;
        }
        else if(n%2==0)
        {
            printf("O NUMERO %d E PAR\n", n);
        }
        else
            printf("O NUMERO %d E IMPAR\n", n);
    }
    while(c!=0);
}