错误：重载函数“ sqrt”的多个实例与参数列表匹配。 那是什么

Question

In my homework for my C class, we have to write a program that checks if an integer is prime. I am getting an error at the sqrt() function. My professor told me that num must be an integer and we must use the sqrt() function. I thought the problem was that the sqrt() function can't be used on an integer but my professor told me that it can, and that I was getting an error from something else. Do you guys see the problem?

int primality(int num)
{

int isprime;

    /*check if num is prime*/
    for (int i = 2; i <= sqrt(num); i++)
    {
        if (num % i == 0)
            isprime = 0; /*is not prime*/
        else 
            isprime = 1; /*is prime*/
    }

    if (isprime == 0)
        return 0;
    elseif (isprime == 1)
        return 1;
}

EDIT: Yes I am using math.h and compiling as C code.

The error msg is "Error: More than one instance of overloaded function "sqrt" matches the argument list.

Answer 1

The C language does not have "overloads". My bet is that you are compiling your code as C++, not C. If you're using GCC, compile with gcc , not g++ . If you are using Visual Studio, there is an option in the properties of the project:

1. right click on your project
2. click properties
3. click on C/C++
4. click on advanced
5. set the property "compile as" to C Code (/TC)

In either case, name your file with a .c extension ( lowercase 'c').

Indeed, in C there is only one sqrt , defined as

double sqrt(double);

and integers convert to double.

Answer 2

You don't want to return 1 inside your loop - it needs to be after the loop completes. Your professor is right that sqrt(num) will work - num will be promoted to double automatically - C has rules for type changes in function calls, etc. You do need to include math.h if you haven't done that elsewhere in your program - what error are you getting?

Answer 3

Your logic is wrong right here:

    if (num % i == 0)
        return 0; /*is not prime*/
    else 
        return 1; /*is prime*/

If you think about it, if the number is not divisible by two, your function immediately returns 1 .

You are return ing too early. Remember that a return essentially stops your function and breaks out of it.

Instead, try to break out only when the number is composite and return True only when you are sure that it is prime.

Just as a side note, your function will register squares of primes (ie 9 , 25 , 49 ) as prime because you are using the < sign in your for loop.

Change it to a <= sign, as that will account for the square root being the only divisor of the number.

Answer 4

The problem is that the sqrt function does not take an integer as a argument. You will want to do

sqrt((double)n) or sqrt((float)n) to fix it.

Answer 5

如果包含#include <math.h> ，则可以通过将sqrt(num)更改为简单的声明变量（例如n, x, i... sqrt(num)来修正以下错误： for (int i = 2; i <= sqrt(num); i++) n, x, i... ），因为sqrt(num)已在math.h库中使用。

Answer 6

the function of sqrt() define as double sqrt(double) so you're input should be double or float.

To turning an integer variable to float without change the type use float( integer_Variable )

For example in you're code write i<=sqrt(float(num));

Answer 7

Did you #include <math.h> ?

Also, according to your function 1, 2, and 3 come back as not prime... Might want to fix that. I'd offer more help but it sounds like this is for homework and I'd rather not just give you the answer (some universities consider it cheating).

Answer 8

要回答您的实际问题：sqrt（）的int没有重载，但double和float都有重载，因此编译器无法猜测要使用哪个重载。