CS2064「術語不考慮帶1個參數的函數」？

Question

我看了其他線程，他們說「這或那不是一個指針!!!」我真的不知道這意味着什麼，我已經聲明瞭所以什麼不在這裏工作？我們正在編寫一個程序來查找一個方形btw的區域，這也不是整個代碼！第2行也是錯誤點。

int s = ((line_one + line_two + line_three)/2); 
    int area = sqrt((s - line_one)(s - line_two)(s - line_three)); 

    cout << "The area of the triange is..." << area << endl; 

此處，我聲明其他變量：

// Area of a Triangle 

double x1 = 0; 
double y1 = 0; 
double x2 = 0; 
double y2 = 0; 
double x3 = 0; 
double y3 = 0; 
double x4 = 0; 
double y4 = 0; 
double x5 = 0; 
double y5 = 0; 
double x6 = 0; 
double y6 = 0; 

//line one points 
cout << "enter point x1" << endl; 
cin >> x1; 
cout << "enter point y1" << endl; 
cin >> y1; 
cout << "enter point x2" << endl; 
cin >> x2; 
cout << "enter point y2" << endl; 
cin >> y2; 
const int line_one = sqrt((pow((x2 - x1), 2) + (pow((y2 - y1), 2)))); 

//line two points 
cout << "enter point x3" << endl; 
cin >> x3; 
cout << "enter point y3" << endl; 
cin >> y3; 
cout << "enter point x4" << endl; 
cin >> x4; 
cout << "enter point y4" << endl; 
cin >> y4; 
const int line_two = sqrt((pow((x4 - x3), 2) + (pow((y4 - y3), 2)))); 

//line three points 
cout << "enter point x5" << endl; 
cin >> x5; 
cout << "enter point y5" << endl; 
cin >> y5; 
cout << "enter point x6" << endl; 
cin >> x6; 
cout << "enter point y6" << endl; 
cin >> y6; 
const int line_three = sqrt((pow((x6 - x5), 2) + (pow((y6 - y5), 2)))); 
//Calculating the Area 

int s = ((line_one + line_two + line_three)/2); 
int area = sqrt((s - line_one)(s - line_two)(s - line_three)); 

    cout << "The area of the triange is..." << area << endl; 


return 0; 
} 

Answer 1

這是一個簡單的錯誤，使你忘了添加*您想在開方（）函數調用乘以事物之間。

int s = ((line_one + line_two + line_three)/2); 
    int area = sqrt((s - line_one)*(s - line_two)*(s - line_three)); 

    cout << "The area of the triange is..." << area << endl; 

就像那樣。它給了你指針錯誤，因爲它認爲你正在試圖調用一個函數，例如（s-line_one）將返回一個函數指針，而（s-line_two）將是該函數的一個參數。