需要帮助来理解传递参数给函数

Question

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I'm new to C++ and unsure about how to pass arguments to functions.

I'm using a function Distance() to calculate the distance between two nodes. I declare the function like this:

int Distance(int x1, int y1, int x2 , int y2)
{   
    int distance_x = x1-x2;
    int distance_y = y1- y2; 
    int distance = sqrt((distance_x * distance_x) + (distance_y * distance_y));
    return distance; 
}

In the main memory I have 2 for loops. What I need to know is if I can pass the values like this: Distance (i, j, i+1, j+1) .

for(int i = 0; i < No_Max; i++)
{
    for(int j = 0; j < No_Max; j++)
    {
        if(Distance(i, j, i+1, j+1) <= Radio_Range) // the function 
            node_degree[i] = node_degree[i] + 1;

        cout << node_degree[i] << endl;
    }  
}

Answer 1

函数的参数可以作为与该参数的类型匹配的任何表达式提供，也可以强制转换为该参数。

Answer 2

Tips : You should use double instead of int if you want to use sqrt :

http://www.cplusplus.com/reference/clibrary/cmath/sqrt/

Answer 3

It looks as if you are calling your Distance(int, int, int, int) function correctly. The following statement will call Distance() :

Distance (i, j, i+1, j+1);

This will store the value returned by Distance() in a variable:

int dist = Distance (i, j, i+1, j+1);

This will compare the value returned by Distance() (the left operand) to Radio_Range (the right operand). If the left operand is less than or equal to the right operand, it will be evaluated to 1 (true). Otherwise it will be 0 (false). If the overall value of the expression inside the if statement is non-zero, the statement or block immediately following the if statement will be executed:

if(Distance(i, j, i+1, j+1) <= Radio_Range)
   //Statement;

or:

if(Distance(i, j, i+1, j+1) <= Radio_Range){
   //Statement;
   //Statement;
   //...
}

However, the value returned by Distance() will be truncated to an integer. Thus, distance will not equal the actual distance unless (distance_x * distance_x) + (distance_y * distance_y) is a perfect square. For better precision, consider using a double . If you intend to have the function return an int , it would be wise to do an explicit type cast, eg:

int distance = (int)sqrt((distance_x * distance_x) + (distance_y * distance_y));

This will ensure that if you or anyone else looks at the code later on, they will not think the function is using the wrong data type.