C语言 指向数组中第一个元素的指针！（C）
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Pointer to first element in array! (C)
提问by Phil
I'm new to C.
我是 C 的新手。
I know this has been asked in many forms but mine is a little unique...I guess. I have an unsigned short pointer.
我知道这已经以多种形式被问过，但我的有点独特......我猜。我有一个无符号的短指针。
6 unsigned short *pt;
7 pt = myArray[0];
The array is declared as such: const unsigned short myArray[1024]and is an array of hex numbers of the form 0x0000 and so on.
该数组声明如下：const unsigned short myArray[1024]并且是 0x0000 等形式的十六进制数数组。
I try to compile, it throws these errors:
我尝试编译，它抛出这些错误：
myLib.c:7: error: data definition has no type or storage class
myLib.c:7: error: type defaults to 'int' in declaration of 'pt'
myLib.c:7: error: conflicting types for 'pt'
myLib.c:6: note: previous declaration of 'pt' was here
myLib.c:7: error: initialization makes integer from pointer without a cast
any ideas of what's going wrong?
关于出了什么问题的任何想法？
Thanks, Phil
谢谢，菲尔
回答by Dietrich Epp
My guess (you only show two lines) is that this code appears outside a function. This is a statement:
我的猜测（你只显示了两行）是这段代码出现在函数之外。这是一个声明：
pt = myArray[0];
Statements must go in functions. Also, if myArrayhas type unsigned short[], then you want to do one of these instead:
语句必须在函数中。此外，如果myArray有 type unsigned short[]，那么您想要执行以下操作之一：
pt = myArray;
pt = &myArray[0]; // same thing
回答by catfish_deluxe_call_me_cd
&is the reference operator. It returns the memory address of the variable it precedes. Pointers store memory addresses. If you want to "store something in a pointer" you dereference it with the *operator. When you do that the computer will look into the memory address your pointer contains, which is suitable for storing your value.
&是参考运算符。它返回它前面的变量的内存地址。指针存储内存地址。如果你想“在一个指针中存储一些东西”，你可以用*操作符取消对它的引用。当您这样做时，计算机将查看您的指针包含的内存地址，该地址适合存储您的值。
char *pc; // pointer to a type char, in this context * means pointer declaration
char letter = 'a'; // a variable and its value
pc = &letter; // get address of letter
// you MUST be sure your pointer "pc" is valid
*pc = 'B'; // change the value at address contained in "pc"
printf("%c\n", letter); // surprise, "letter" is no longer 'a' but 'B'
When you use myArray[0]you don't get an address but a value, that's why people used &myArray[0].
当您使用时，myArray[0]您得到的不是地址而是值，这就是人们使用&myArray[0].
回答by Jonathan Wood
Yeah, you really should include a bit more code so we can see the context.
是的，您确实应该包含更多代码，以便我们可以查看上下文。
I don't quite get the error messages, but your code is not correct.
我不太明白错误消息，但您的代码不正确。
Try:
尝试：
pt = &myArray[0];
Or:
或者：
pt = myArray + 0;
Or just:
要不就：
pt = myArray;
Instead.
反而。
回答by David Grayson
You should not throw away the const qualifier; so the pointer should have a const modifier.
你不应该扔掉 const 限定符；所以指针应该有一个 const 修饰符。
const unsigned short myArray[1024];
const unsigned short * pointer = myArray; // set pointer to first element
