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Why const char * and const char [] as function parameter are considered to be equivalent? (2 answers)

When a function has a specific-size array parameter, why is it replaced with a pointer? (3 answers)

What is array-to-pointer conversion aka. decay? (11 answers)

Closed 19 mins ago.

请允许我向我解释这些职能是否相同?

void doubleArray(int* values, int length)

void doubleArray(int values[], int length)

我不得不改变我的法典,以完全相同,但我不相信逻辑。我预计至少必须改变职务中写的内容。如果需要的话,全面法典如下:

  #include <iostream>

    using std::cout;
    using std::endl;

void doubleArray(int* values, int length);

int main() {

    int length = 10;
    int values[length];

    for (int i = 0; i < length; i++){
        values[i] = i;
    }

    doubleArray(values, length);

    for (int i = 0; i < length; i++) {
      cout << values[i] << endl;
   }

}

void doubleArray(int* values, int length) {
    for(int i = 0; i < length; i++){
        values[i] = values[i] * 2;
    }
}

Answer 1

From https://en.cppreference.com/w/cpp/language/function#Parameter_list:

The type of each function parameter in the parameter list is determined according to the following rules:

[。]

2) If the type is "array of T" or "array of unknown bound of T", it is replaced by the type "pointer to T"

[。]

由于这些规则,以下职能声明完全相同:
int f(char[]);
int f(char* s);
int f(char* const);
int f(char* volatile s);

Answer 2

他们为什么也一样? 由于这符合C++(从C继承)的规则。

C的设计者认为,能够把一个阵列变成一个功能,是一种坏的想法。他们认为,最好是对阵列的第一点进行点击。问题(我认为)。

但是,他们还决定,他们不想在功能参数申报中放弃阵列。因此,他们决定,当作为功能参数使用时,阵列表。 T param[] is only a different way of written the pointer form, T* param. 如果你指明阵列的大小(被忽略),那么,T param [10]<>> > 代码也是个点。更值得怀疑(我认为)。

在C++中,通过<代码>std:vector或std:array,你能够避免这一切。在C,你不是这样幸运的。

Answer 3

附录:

int values[10];

分配给10个帐篷的一组记忆被分配在盒子上,可变的<条码>价值<> 代码>包含这种记忆阵列开始的地址,或以另一种方式处理阵列零件,即相同。

因此,

&values[0] == values

总是真实的。

同样,这两项声明也涉及同样的内容。

cout << values[5];
cout << *(values + 5);

因此, if we have a function parameter (or any other variable), it can be declared as of type int* or int[] and both will accept values, because both declarations are merely alternative ways of declaring the same thing - a pointer. .

这可能会造成混乱,因为这两项要求都有效:

void foo (int[] vals);

int values[10];
foo (values); // Passing an array

int x = 0;
int* px = &x:
foo (px);     // Passing a single object by address

So as a matter of style, if a function expects an array, it should take the array for (int[]) and a length. If it expects a single object it should take a pointer (int*).

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