在这种情况下， A [4] 是 4 数组中的整数 A ， AP 是整数的指针，所以你分配给一个整数指针，这就是警告。照片所以 AP 现在持有 45 ，当你试图去参考它（通过执行 * AP ）你想在地址45，这是一个无效的地址来访问内存，所以你的程序崩溃。

In this case a[4] is the 4th integer in the array a, ap is a pointer to integer, so you are assigning an integer to a pointer and that's the warning.
So ap now holds 45 and when you try to de-reference it (by doing *ap) you are trying to access a memory at address 45, which is an invalid address, so your program crashes.

您应该做的 AP =放大器（A [4]）; 或 AP = A + 4;

在 C 数组名衰变为指针，所以 A 指向数组的第一个元素。
这样， A 等同于＆功放;（A [0]）

In c array names decays to pointer, so a points to the 1st element of the array.
In this way, a is equivalent to &(a[0]).