tl;dr:当您使用已定义的数组时，实际上是在使用指向其第一个元素的指针。

因此:

当你写arr[idx]时，你实际上是在说*(arr + idx)。 函数从来不会真正将数组作为参数，只接受指针——当指定数组参数时可以直接接受，如果将引用传递给数组则可以间接接受。

这条规则的例外情况:

可以将固定长度的数组传递给结构中的函数。 Sizeof()给出数组占用的大小，而不是指针的大小。

数组衰减意味着，当数组作为参数传递给函数时，它被视为(“衰减为”)指针。

void do_something(int *array) {
  // We don't know how big array is here, because it's decayed to a pointer.
  printf("%i\n", sizeof(array));  // always prints 4 on a 32-bit machine
}

int main (int argc, char **argv) {
    int a[10];
    int b[20];
    int *c;
    printf("%zu\n", sizeof(a)); //prints 40 on a 32-bit machine
    printf("%zu\n", sizeof(b)); //prints 80 on a 32-bit machine
    printf("%zu\n", sizeof(c)); //prints 4 on a 32-bit machine
    do_something(a);
    do_something(b);
    do_something(c);
}

上述情况有两个复杂情况或例外情况。

首先，在C和c++中处理多维数组时，只丢失了第一个维度。这是因为数组在内存中是连续布局的，所以编译器必须知道除第一个维度以外的所有维度，才能计算该内存块的偏移量。

void do_something(int array[][10])
{
    // We don't know how big the first dimension is.
}

int main(int argc, char *argv[]) {
    int a[5][10];
    int b[20][10];
    do_something(a);
    do_something(b);
    return 0;
}

其次，在c++中，您可以使用模板来推断数组的大小。微软将此用于c++版本的Secure CRT函数(如strcpy_s)，您可以使用类似的技巧可靠地获取数组中的元素数量。

数组与C/ c++中的指针基本相同，但又不完全相同。一旦你转换一个数组:

const int a[] = { 2, 3, 5, 7, 11 };

转换为指针(不需要强制转换，因此在某些情况下可能会意外发生):

const int* p = a;

你将失去sizeof操作符计算数组中元素的能力:

assert( sizeof(p) != sizeof(a) );  // sizes are not equal

这种丧失的能力被称为“衰退”。

有关更多详细信息，请参阅这篇关于数组衰减的文章。

tl;dr:当您使用已定义的数组时，实际上是在使用指向其第一个元素的指针。

因此:

当你写arr[idx]时，你实际上是在说*(arr + idx)。 函数从来不会真正将数组作为参数，只接受指针——当指定数组参数时可以直接接受，如果将引用传递给数组则可以间接接受。

这条规则的例外情况:

可以将固定长度的数组传递给结构中的函数。 Sizeof()给出数组占用的大小，而不是指针的大小。

试试这段代码

void f(double a[10]) {
    printf("in function: %d", sizeof(a));
    printf("pointer size: %d\n", sizeof(double *));
}

int main() {
    double a[10];
    printf("in main: %d", sizeof(a));
    f(a);
}

你会看到函数中数组的大小并不等于main中数组的大小，而是等于指针的大小。

你可能听说过“数组是指针”，但是，这不是完全正确的(sizeof在main打印正确的大小)。然而，当传递时，数组衰减为指针。也就是说，不管语法显示的是什么，实际上传递了一个指针，函数实际上接收了一个指针。

在这种情况下，定义void f(double a[10]被编译器隐式转换为void f(double *a)。你可以直接将函数参数声明为*a。您甚至可以编写一个[100]或[1]，而不是[10]，因为它实际上从未以这种方式编译(但是，您显然不应该这样做，这会使读者感到困惑)。

数组是由c语言中的指针自动传递的，其背后的原理只能推测。

int a[5], int *a and int (*a)[5] are all glorified addresses meaning that the compiler treats arithmetic and deference operators on them differently depending on the type, so when they refer to the same address they are not treated the same by the compiler. int a[5] is different to the other 2 in that the address is implicit and does not manifest on the stack or the executable as part of the array itself, it is only used by the compiler to resolve certain arithmetic operations, like taking its address or pointer arithmetic. int a[5] is therefore an array as well as an implicit address, but as soon as you talk about the address itself and place it on the stack, the address itself is no longer an array, and can only be a pointer to an array or a decayed array i.e. a pointer to the first member of the array.

例如，在int (*a)[5]上，对a的第一次解引用将产生int *(因此相同的地址，只是不同的类型，注意不是int a[5])，对a的指针算术，即a+1或*(a+1)将根据5个int数组的大小(这是它所指向的数据类型)，第二次解引用将产生int。然而，对于int类型a[5]，第一次解引用将产生int类型，指针的算术将与int类型的大小相关。

To a function, you can only pass int * and int (*)[5], and the function casts it to whatever the parameter type is, so within the function you have a choice whether to treat an address that is being passed as a decayed array or a pointer to an array (where the function has to specify the size of the array being passed). If you pass a to a function and a is defined int a[5], then as a resolves to an address, you are passing an address, and an address can only be a pointer type. In the function, the parameter it accesses is then an address on the stack or in a register, which can only be a pointer type and not an array type -- this is because it's an actual address on the stack and is therefore clearly not the array itself.

You lose the size of the array because the type of the parameter, being an address, is a pointer and not an array, which does not have an array size, as can be seen when using sizeof, which works on the type of the value being passed to it. The parameter type int a[5] instead of int *a is allowed but is treated as int * instead of disallowing it outright, though it should be disallowed, because it is misleading, because it makes you think that the size information can be used, but you can only do this by casting it to int (*a)[5], and of course, the function has to specify the size of the array because there is no way to pass the size of the array because the size of the array needs to be a compile-time constant.