c while循环在条件之前停止

Question

I have one exercise that I have to calculate fibonacci numbers until 100 and then print them.

I have made this code:

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

    int main()
    {

        int i=2,fibonacci = 0, fParcialone = 1, fParcialtwo = 0;
        printf("The %2dst Fibonacci number is %3d\n", i-1, fibonacci+1);
        while (fibonacci <= 100){

            fibonacci = fParcialone+fParcialtwo;
            printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
            fParcialtwo = fParcialone;
            fParcialone = fibonacci;
            i++;
        }
        return 0;
    }

As you can see, it prints the 12th number and this number is greater than 100. I understand why it is doing that.

One alternative is to make this:

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

    int main()
    {

        int i=2,fibonacci = 0, fParcialone = 1, fParcialtwo = 0;
        printf("The %2dst Fibonacci number is %3d\n", i-1, fibonacci+1);
        while (fibonacci <= 100){

            fibonacci = fParcialone+fParcialtwo;
            if (fibonacci > 100){
                return 0;
            }
            printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
            fParcialtwo = fParcialone;
            fParcialone = fibonacci;
            i++;
        }
        return 0;
    }

And now it works but now, In every loop it as to make two comparisons and not just one and I believe that this way it uses to much "processor time" (in this example its to little but on a bigger scale it probably will make difference).

Is there a better way to do this?

favolas

Answer 1

I think that you can use a do ... while(cond). It should looks like :

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

int main()
{

    int i=2,fibonacci = 0, fParcialone = 1, fParcialtwo = 0;
    printf("The %2dst Fibonacci number is %3d\n", i-1, fibonacci+1);
    fibonacci = fParcialone+fParcialtwo;
    do {
        printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
        fParcialtwo = fParcialone;
        fParcialone = fibonacci;
        i++;
    } while ((fibonacci = fParcialone+fParcialtwo) <= 100);
    return 0;
}

Answer 2

The processing time of one simple if-statement is quite negligible.

If you insist on making the code better looking one option is simply replacing the while condition with:

while (fParcialone + fParcialtwo <= 100)

or change the loop to a do-while loop.

Answer 3

The simplest solution is:

    ...
    while (1) {

        fibonacci = fParcialone+fParcialtwo;
        if (fibonacci > 100){
            return 0;
        }
        printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
        fParcialtwo = fParcialone;
        fParcialone = fibonacci;
        i++;
    }
    ...

Answer 4

you can calculate the first number outside of loop ,and each time calculate it in the end of the loop. like this:

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

    int main()
    {

        int i=2,fibonacci = 0, fParcialone = 1, fParcialtwo = 0;
        printf("The %2dst Fibonacci number is %3d\n", i-1, fibonacci+1);

        fibonacci = fParcialone+fParcialtwo;
        while (fibonacci <= 100){

            printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
            fParcialtwo = fParcialone;
            fParcialone = fibonacci;
            i++;
            fibonacci = fParcialone+fParcialtwo;
        }
        return 0;
    }

Answer 5

My C is a couple of decades rusty so I won't attempt valid code, but to solve this problem you basically need to break out of the loop in the middle to remove the test.

while(true)
{
   calculate
   if (fib <= 100)
   {
       output
   }
   else
   {
       break; // Exit from the loop
   }
}

However - in truth, the extra comparison just isn't an issue in terms of performance. Even at scale the hit should be vanishingly small in comparison with other operations so really you have far more important things to worry about in the general case. (There will always be specific cases where that might not be true.)

Answer 6

Sure there is.

When you run into something like this, if you feel that you need to put a condition inside the loop like that, it can probably go in the conditional check that happens in the while statement and whatever follows after your inner conditional is what should be at the very start of the while loop. Then just rearrange everything as needed (like on a circle) and add initialization statements as needed before the loop to get it started.

int main()
{

    int i=2,fibonacci = 0, fParcialone = 1, fParcialtwo = 0;
    printf("The %2dst Fibonacci number is %3d\n", i-1, fibonacci+1);
    //initialize
    fibonacci = fParcialone+fParcialtwo;
    while (fibonacci <= 100){
        printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
        fParcialtwo = fParcialone;
        fParcialone = fibonacci;
        fibonacci = fParcialone+fParcialtwo;
        i++;
    }
    return 0;
}

Answer 7

If you want to keep most of the structure of your code, try :

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

int main()
{

    int i=1,fibonacci = 1, fParcialone = 1, fParcialtwo = 0;
    while (fibonacci <= 100){
        printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
        fibonacci = fParcialone+fParcialtwo;
        fParcialtwo = fParcialone;
        fParcialone = fibonacci;
        i++;
    }
    return 0;
}

Answer 8

We already know the first Fibonacci number is 1 so you don't even need to calculate it before the condition statement. I know it is different code than yours, was just having some fun, but hopefully it gets the idea across.

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

int Fibonacci(int n)
{
    if(n <= 1) return n;  
    return Fibonacci(n-2) + Fibonacci(n-1);
}

int main()
{
    int i = 1, fibonacci = 1;
    do 
    {
        printf("The %2dst Fibonacci number is %3d\n", i, fibonacci);
        fibonacci = Fibonacci(++i);
    } while (fibonacci <= 100);

    return 0;
};

The key is calculating and checking the 12th Fibonacci number in this case, before printing it to the screen. Either way, one if-check won't have any noticeable processing time and it is negligible.