C++ 错误：一元“*”的无效类型参数

error: invalid type argument of unary '*'

提问by user3502479

I don't understand these errors can someone explain?

我不明白这些错误有人可以解释吗？

error: invalid type argument of unary '' (have 'double') error: invalid type argument of unary '' (have 'double') error: invalid type argument of unary '*' (have 'double')

错误：一元' '的无效类型参数（有'double'）错误：一元''的无效类型参数（有'double'）错误：一元'*'的无效类型参数（有'double'）

    double getMedian(double *array, int *hours){
    if (*hours <= 0) return 0;
    if (*hours % 2) return (float)*array[(*hours + 1) / 2];
    else{int pos = *hours / 2;
    return (float)(*array[pos] + *array[pos + 1]) / 2;}}

回答by ppl

You are already dereferencing arraywith the []operator. What you want is:

您已经在取消array对[]运算符的引用。你想要的是：

double getMedian(double *array, int *hours){
if (*hours <= 0) return 0;
if (*hours % 2) return (float)array[(*hours + 1) / 2];
else{int pos = *hours / 2;
return (float)(array[pos] + array[pos + 1]) / 2;}}

Note that writing x[y]is shorthand for *(x + (y)). In your code, you have essentially have the equivalent of **array.

请注意，写作x[y]是 的简写*(x + (y))。在您的代码中，您基本上拥有相当于**array.

回答by sajas

When you use the [] operator on the arrays or pointers, you don't have to dereference them again to get the value. you could just say,

当您在数组或指针上使用 [] 运算符时，您不必再次取消引用它们来获取值。你只能说，

if (*hours % 2) return (float)array[(*hours + 1) / 2];

and

和

return (float)(array[pos] + (array[pos + 1]) / 2);

回答by John3136

*array[(*hours + 1) / 2];so arrayis an array of doubles. You treating it as a 2D array because you try to dereference once via *and one via [].

*array[(*hours + 1) / 2];array双打数组也是如此。您将其视为二维数组，因为您尝试取消引用一次 via*和一次 via []。

Also, I'd add some ()to all of that to make it a bit clearer without having to memorise the order of operations.

此外，我会()在所有这些内容中添加一些内容，以便在不必记住操作顺序的情况下更清晰一些。