C语言 C 指针和数组：[警告] 赋值使指针来自整数而不进行强制转换
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C pointers and arrays: [Warning] assignment makes pointer from integer without a cast
提问by user2274889
I'm having some trouble with pointers and arrays in C. Here's the code:
我在使用 C 中的指针和数组时遇到了一些麻烦。代码如下：
#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;
}
When I compile and run the code I get this warning:
当我编译并运行代码时，我收到此警告：
[Warning] assignment makes pointer from integer without a cast [enabled by default]
[警告] 赋值使指针从整数而不进行强制转换 [默认启用]
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.
对于第 9 行（ap = a[4];），终端崩溃。如果我将第 9 行更改为不包含位置 (ap = a;)，则不会收到任何警告并且可以正常工作。为什么会这样？我觉得答案很明显，但我就是看不到。
回答by Dipto
In this case a[4]is the 5thinteger in the array a, apis a pointer to integer, so you are assigning an integer to a pointer and that's the warning.
So apnow holds 45and 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.
在这种情况下a[4]是5th数组中的整数a，ap是指向整数的指针，因此您将整数分配给指针，这就是警告。
所以ap现在成立45，当您尝试取消引用它（通过执行*ap）时，您正在尝试访问地址 45 处的内存，这是一个无效地址，因此您的程序崩溃。
You should do ap = &(a[4]);or ap = a + 4;
你应该做ap = &(a[4]);或ap = a + 4;
In carray names decays to pointer, so apoints to the 1st element of the array.
In this way, ais equivalent to &(a[0]).
在c数组中，名称衰减为指针，因此a指向数组的第一个元素。
这样，a等价于&(a[0])。
回答by halfbit
What are you doing: (I am using bytes instead of in for better reading)
你在做什么：（为了更好地阅读，我使用字节而不是 in）
You start with int *apand so on, so your (your computers) memory looks like this:
你从int *ap等等开始，所以你的（你的计算机）内存看起来像这样：
-------------- memory used by some one else --------
000: ?
001: ?
...
098: ?
099: ?
-------------- your memory --------
100: something <- here is *ap
101: 41 <- here starts a[]
102: 42
103: 43
104: 44
105: 45
106: something <- here waits x
lets take a look waht happens when (print short cut for ...print("$d", ...)
让我们看看什么时候发生 (print short cut for ...print("$d", ...)
print a[0] -> 41 //no surprise
print a -> 101 // because a points to the start of the array
print *a -> 41 // again the first element of array
print a+1 -> guess? 102
print *(a+1) -> whats behind 102? 42 (we all love this number)
and so on, so a[0] is the same as *a, a[1] = *(a+1), ....
依此类推，所以 a[0] 与 *a 相同，a[1] = *(a+1), ....
a[n] just reads easier.
a[n] 只是读起来更容易。
now, what happens at line 9?
现在，第 9 行会发生什么？
ap=a[4] // we know a[4]=*(a+4) somehow *105 ==> 45
// warning! converting int to pointer!
-------------- your memory --------
100: 45 <- here is *ap now 45
x = *ap; // wow ap is 45 -> where is 45 pointing to?
-------------- memory used by some one else --------
bang! // dont touch neighbours garden
So the "warning" is not just a warning it's a severe error.
所以“警告”不仅仅是一个警告，它是一个严重的错误。
回答by njzk2
int[]and int*are represented the same way, except int[] allocates (IIRC).
int[]并且int*以相同的方式表示，除了 int[] 分配（IIRC）。
apis a pointer, therefore giving it the value of an integer is dangerous, as you have no idea what's at address 45.
ap是一个指针，因此给它一个整数值是危险的，因为你不知道地址 45 是什么。
when you try to access it (x = *ap), you try to access address 45, which causes the crash, as it probably is not a part of the memory you can access.
当您尝试访问它 ( x = *ap) 时，您尝试访问地址 45，这会导致崩溃，因为它可能不是您可以访问的内存的一部分。
