Question

我刚刚在“C编程语言”一书中读到，数组不是变量，并且看到数组到指针的分配（反之亦然）不能作为结果。那么如果数组不是变量那么它是什么？

Answer 1

Array是一个包含许多值的数据结构，所有值都具有相同的类型，并且数组名称是不可修改的 l值（命名的内存位置） - 可以寻址，但不可修改。这意味着它不能被修改或不能成为赋值运算符的左操作数。

int a[10] = {0};
int *p = a;     //OK
a++             // Wrong
a = p;          // Wrong

Answer 2

int numbers[] = {1,2,3}

numbers不是变量，它是数组名称，它只是数组中第一个元素的地址。要验证这一点，请执行以下操作，查看numbers的地址和numbers[0]的地址：printf("%p and %p and %p", &numbers, numbers, &numbers[0]);所有三个指针都具有相同的值，因为numbers是什么都没有但是数组中第一个元素的地址。因此numbers不是包含指针或值的变量，因为它在内存中没有专用地址来存储值。

但是，请查看此指针变量：

int *pnumbers = numbers;
`printf("%p and %p and %p", &pnumbers, pnumbers, &pnumbers[0]);`

您会注意到&pnumbers在内存中有不同的地址，因为pnumber在内存中有一个专用地址，用于存储数组中第一个元素的地址numbers。

将所有代码放在一起：

#include <stdio.h>

main(){
    int numbers[] = {1,2,3};
    printf("%p and %p and %p\n", &numbers, numbers, &numbers[0]); // Address of numbers, value of numbers, first element of numbers

    int *pnumbers = numbers;
    printf("%p and %p and %p\n", &pnumbers, pnumbers, &pnumbers[0]); // Address of pnumbers, value of pnumbers, first element of the array pnumbers is pointing to
}

<强>输出

0xbfb99fe4 and 0xbfb99fe4 and 0xbfb99fe4 // All three have the same address which is the address of the first element in the array
0xbfb99fe0 and 0xbfb99fe4 and 0xbfb99fe4 // The first one is different since pnumbers has been allocated a memory address to store a pointer which is the first element of the array numbers

Answer 3

这是一个占位符。用于表示引用存储器的顺序部分的常用方法的符号。它本身并不是变量。

int main(int argc, char** argv)
{
  int array[10];
  int value'
  int* pointer;

  value = array[0];   // Just fine, value and array[0] are variables
  array[0] = value;   // Just fine, value and array[0] are variables
  pointer = &array[0];   // Just fine, &array[0] is an address
  pointer = array;    // Also just fine 
  //Because the compiler treats "array" all by itself as the address of array[0]  
  //That is: array == &array[0]

  &array[0] = pointer   // ERROR,  you can't assign the address of something to something else.
  array = pointer;      // ERROR, array is not a variable, and cannot be assigned a value.

  //Also bad, but technically they compile and could theoretically have their use
  pointer = value;
  pointer = array[0];
  array[0] = pointer; 
  //Intermixing pointers and non-pointer variables is generally a bad idea.
}

array通常被视为变量，因为它表示该内存块的（第一项）的地址。但这不是变数。它没有自己的内存来存储任何东西。人们将指针设置为“数组”，因为它是一个方便的约定，编译器知道这意味着什么，而且它很常见。

Answer 4

数组是相同类型的元素序列。如果指定一个指向数组的指针，指针将指向数组中第一个变量的地址。

int a[10];
int *p;
int *p2;
p = a;
p2 = &a[0];
printf("%d\n", p == p2);

输出：

Answer 5

 I've just read in the book "The C Programming Language" that array is not a 
 variable and saw that assignment of array to pointers (and vice versa)

允许反之亦然...数组就像一个常量指针（你不能改变它指向的地址）。但是，您可以将该地址分配给指针。

 #include <stdio.h>

 int main()
 {
    int x[3] = {1, 2, 3};

    int *p = x;
    p[0] = 50;
    printf("%d", x[0]);
 }

C中定义的数组类型

5 个答案: