Even Number Counting (c language)

Question

I need to display on the screen the numbers between a typed initial and final value, after which write how many are pairs. ex. 2 and 10 will display on screen: 2, 3, 4, 5, 6, 7 ,8 ,9, 10.

then tell the Qtd of pairs have there (would be 5, in case). I can’t perform that last part.

for (val1; val1 <= val2; val1++)
    {
          printf("\n%d ", val1);
    }
    if (val1%2==0)
    {
    par++;
    }
    printf ("\n Entre os valores %d e %d, existem %d números pares.", val1, val2, par);

Answer 1

The problem is you’re not comparing whether the val1 is even or not within the loop for, you are "closing" the loop before that:

for (val1; val1 <= val2; val1++)
    {
          printf("\n%d ", val1);
>>> }
    if (val1%2==0)
...

Just remove the } which is just after this printf and add it at the end (after the final key of the if):

...
for (val1; val1 <= val2; val1++)
    {   
    printf("\n%d ", val1);

    if (val1 % 2 == 0)
    {   
            par++;
    }   
}
printf ("\n Entre os valores %d e %d, existem %d números pares.", val1-val2, val2, par);
...

This way the code will work as desired.

Answer 2

Like looks like an exercise, it may be "mandatory" to iterate one by one and use the operator % (and there the another answer I’ve told you how to do).

But you can do a little better. Just make sure you’re starting with an even number, and then just iterate from 2 to 2:

// ... ler val1 e val2

int n = val1;
// se o valor inicial é ímpar, soma 1 para que ele seja par
if (n % 2 != 0)
    n++;

// agora eu sei que n é par, então posso fazer o for de 2 em 2
int cont = 0;
for(; n <= val2; n += 2) {
    printf("%d\n", n);
    cont++;
}
printf("Entre os valores %d e %d existem %d números pares.\n", val1, val2, cont);

That is, if val1 is odd, I add 1 so that the for start at an even number. And as I have now ensured that it always starts at an even number, I can do the for by 2 (i.e., n += 2). So I just use the operator % once, before the loop (instead of iterating through all the numbers, one by one, and using % for all).

And see that in the first expression inside the for you don’t need to put val1 "loose" there, can leave empty even (if you are not going to do anything, do not put anything).

How do you want to print val1 in the end, so you can’t change it in the loop, so I used another variable for this (in this case, the n).

---

Although, if it was just to know the amount, you don’t even have to keep counting. After all, come on a and b (provided that a be even), the amount of even numbers is ((b - a) / 2) + 1. The loop is only necessary to print the numbers. That is:

int n = val1;
// se o valor inicial é ímpar, soma 1 para que ele seja par
if (n % 2 != 0)
    n++;

// não precisa contar dentro do loop
int cont = ((val2 - n) / 2) + 1;

// agora eu sei que n é par, então posso fazer o for de 2 em 2
for(; n <= val2; n += 2) {
    printf("%d\n", n);
}
printf("Entre os valores %d e %d existem %d números pares.\n", val1, val2, cont);