[英]Calculating distance: method “must return a value”?
我正在嘗試調用dist()
方法,但是我不斷收到錯誤消息,說dist()
必須返回一個值。
// creating array of cities
double x[] = {21.0,12.0,15.0,3.0,7.0,30.0};
double y[] = {17.0,10.0,4.0,2.0,3.0,1.0};
// distance function - C = sqrt of A squared + B squared
double dist(int c1, int c2) {
z = sqrt ((x[c1] - x[c2] * x[c1] - x[c2]) + (y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
}
void main()
{
int a[] = {1, 2, 3, 4, 5, 6};
execute(a, 0, sizeof(a)/sizeof(int));
int x;
printf("Type in a number \n");
scanf("%d", &x);
int y;
printf("Type in a number \n");
scanf("%d", &y);
dist (x,y);
}
將返回類型更改為void:
void dist(int c1, int c2) {
z = sqrt ((x[c1] - x[c2] * x[c1] - x[c2]) +
(y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
}
或在函數末尾返回值:
double dist(int c1, int c2) {
z = sqrt ((x[c1] - x[c2] * x[c1] - x[c2]) +
(y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
return z;
}
dist
函數聲明為返回double
但不返回任何內容。 您需要顯式返回z
或將返回類型更改為void
// Option #1
double dist(int c1, int c2) {
z = sqrt (
(x[c1] - x[c2] * x[c1] - x[c2]) + (y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
return z;
}
// Option #2
void dist(int c1, int c2) {
z = sqrt (
(x[c1] - x[c2] * x[c1] - x[c2]) + (y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
}
您正在將“結果為z”輸出到STDOUT,但實際上沒有作為dist
函數的結果返回它。
所以
double dist(int c1, int c2) {
z = sqrt (
(x[c1] - x[c2] * x[c1] - x[c2]) + (y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
}
應該
double dist(int c1, int c2) {
z = sqrt (
(x[c1] - x[c2] * x[c1] - x[c2]) + (y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
return(z);
}
(假設您仍然要打印它)。
或者
您可以使用void
聲明dist
不返回值:
void dist(int c1, int c2) {
z = sqrt (
(x[c1] - x[c2] * x[c1] - x[c2]) + (y[c1] - y[c2] * y[c1] - y[c2]));
cout << "The result is " << z;
}
請參閱: C ++函數教程 。
只需添加以下行:return z; -1這樣的問題。
由於您已定義dist返回double(“ double dist”),因此在dist()的底部應執行“ return dist;”。 或將“ double dist”更改為“ void dist”-void表示它不需要返回任何內容。
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