C指针和数组：【警告】赋值时将整数指针，未作投 [英] C pointers and arrays: [Warning] assignment makes pointer from integer without a cast

问题描述

我有一些麻烦，在C.指针和数组这里是code：

I'm having some trouble with pointers and arrays in C. Here's the code:

#include<stdio.h>


int *ap;
int a[5]={41,42,43,44,45};
int x;

int main()
{
    ap = a[4];
    x = *ap;
    printf("%d",x);
    return 0;
}

当我编译和运行code我得到这样的警告：

When I compile and run the code I get this warning:

【警告】赋值时将整数指针，未作投
  [默认启用]

[Warning] assignment makes pointer from integer without a cast [enabled by default]

有关行号9（AP =一个[4]）和终端崩溃。如果我改变9号线不包括位置（AP =一;）我没有得到任何警告和它的作品。这究竟是为什么？我觉得答案是显而易见的，但我不能看到它。

For line number 9 (ap = a[4];) and the terminal crashes. If I change line 9 to not include a position (ap = a;) I don't get any warnings and it works. Why is this happening? I feel like the answer is obvious but I just can't see it.

推荐答案

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]).

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