Prime factors in C | 易学教程

问题

I came across this effective program for printing all the prime factors of a given number:

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

// A function to print all prime factors of a given number n
void primeFactors(int n)
{
    // Print the number of 2s that divide n
    while (n%2 == 0)
    {
        printf("%d ", 2);
        n = n/2;
    }

    // n must be odd at this point.  So we can skip 
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i+2)
    {
        // While i divides n, print i and divide n
        while (n%i == 0)
        {
            printf("%d ", i);
            n = n/i;
        }
    }

    // This condition is to handle the case when n 
    // is a prime number greater than 2
    if (n > 2&&n%2!=0)
        printf ("%d ", n);
}

/* Driver program to test above function */
int main()
{
    int n = 315;
    primeFactors(n);
    return 0;
}

The output is 3 3 5 7. That's perfect.

But I have a confusion in this algorithm. The sqrt() returns a float value. If it is displayed in integer format in printf, it returns some random, large number. If this is the case, the condition in the for loop should fail, because sqrt() as an integer returns a random number. Could someone explain this?

Plus,could someone verify this? This algorithm was wrongly written in geeksforgeeks website as if(n>2) instead of if(n>2&&n!=0), which was added by me. Someone please verify this algorithm. Thanks in advance.

回答1:

If you try to print the value of sqrt(n) as if it were an integer:

printf("sqrt(n) = %d\n", sqrt(n)); // Don't do this

then you have undefined behavior. sqrt() returns a result of type double. The compiler doesn't know (or isn't required to know) the expected type of the argument. It's up to you to pass an argument of the correct type. The above call to printf has undefined behavior.

In other contexts, where the type and expected type of an expression are unambiguous, the language requires implicit conversions. In particular, in the expression i <= sqrt(n), where i and n are of type int, the argument n is converted from int to double (the parameter type for sqrt()), and the value ofiis also converted frominttodouble`. The result is likely to be what you expected it to be.

Incidentally, this:

for (int i = 3; i <= sqrt(n); i = i+2) ...

is likely to be inefficient. The sqrt function is relatively expensive, and it's going to be called on each iteration of the loop. (Precomputing sqrt(n) is a good solution in some cases, but here the value of n can change within the loop.)

An alternative is:

for (int i = 3; i*i <= n; i += 2) ...

which avoids the complications of floating-point arithmetic (but think about whether i*i can overflow).

(It would be nice if C had an integer square root function in the standard library, but it doesn't.)

回答2:

As far as I understand from your question that you are passing float to printf() function with %d specifier which is for integer. It's undefined behavior.

N1570 7.21.6.1 paragraph 9:

... If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

That's why you see the garbage value. The compiler may warn you but not error either compile time or run-time since C is not strictly typed language.

It's compiled successfully but its output is -376018152. So again, it's UB.

回答3:

Others have suggested pre-calculating the integer square root, but you need to be careful to recalculate it as needed. For the main loop I would use something like:

// n must be odd at this point.  So we can skip 
// one element (Note i = i +2)
int limit = isqrt(n);  // Calculate integer square root.
for (int i = 3; i <= limit; i = i+2)
{
    // While i divides n, print i and divide n
    if (n % i == 0)
    {
        printf("%d ", i);
        n = n / i;

        while (n%i == 0)
        {
            printf("%d ", i);
            n = n / i;
        }

        // Recalculate limit as n has changed.
        limit = isqrt(n);
    }
}

This only recalculates the square root when the number being tested has changed. I use isqrt(n) to indicate the function to do the actual calculation. In my own code I use Newton-Raphson, but there are other possible methods.

Your test at the end could be simplified to:

// This condition is to handle the case when n 
// is a prime number greater than 2
if (n > 1)
{
    printf ("%d ", n);
}

There is at most one prime factor greater than the square root of a number. That lone factor (if present) will be the remainder after all the smaller prime factors have been divided out.

来源：https://stackoverflow.com/questions/50768926/prime-factors-in-c

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primes

algebra

sqrt