指向数组的c / c ++指针和指向指针的指针 [英] c/c++ pointer to an array vs pointer to a pointer

问题描述

我认为数组和指针基本上是一样的，直到我运行这个程序：

  int main（）{
 int * a = new int（19）; 
 int b [1]; 
 b [0] = 19; 
 
 printf（* a：％d \\\
 a：％p \\\
& a：％p \\\
，* a，a，& a）; 
 printf（* b：％d \\\
 b：％p \\\
& b：％p \\\
，* b，b，& b）; 
 delete a; 
} 
  

输出为：

  * a：19 
a：0x7f94524000e0 
& a：0x7fff51b71ab8 
 * b：19 
b：0x7fff51b71ab4 
& b：0x7fff51b71ab4 
  

有人可以解释为什么& b的输出与b相同？ / p>

谢谢！
-Erben

解决方案

数组和指针是不一样的。指针可以像一个数组（例如通过索引访问）。

& b 是指向整个数组的指针， b 是指向第一个元素的指针。他们可能指向内存中的同一个地址，但它们完全不同。

  + --------- ---------------------- + 
 | + ----- + ----- + ----- + --- -  + ----- + | 
 || | | | | || 
& b ----> || 0 | 1 | 2 | ... | N || 
 || | | | | || 
 | + ----- + ----- + ----- + ----- + ----- + | 
 + --- ^ --------------------------- + 
 | 
 b 
  

I thought array and pointer are basically the same thing, until I run this program:

int main() {
  int* a = new int(19);
  int b[1];
  b[0] = 19;

  printf("*a: %d\n a: %p\n &a:%p\n", *a, a, &a);
  printf("*b: %d\n b: %p\n &b:%p\n", *b, b, &b);
  delete a;
}

output is:

*a: 19
 a: 0x7f94524000e0
 &a:0x7fff51b71ab8
*b: 19
 b: 0x7fff51b71ab4
 &b:0x7fff51b71ab4

can someone please explain why the output of &b is the same as b?

Thanks! -Erben

解决方案

Arrays and pointers are not the same. A pointer can behave like an array (e.g. accessing by index).

&b is a pointer to the whole array and b is a pointer to the first element. They may point to a same address in memory but they are totally different things.

        +-------------------------------+  
        |+-----+-----+-----+-----+-----+|  
        ||     |     |     |     |     ||  
 &b---->||  0  |  1  |  2  | ... |  N  ||  
        ||     |     |     |     |     ||  
        |+-----+-----+-----+-----+-----+|  
        +---^---------------------------+  
            |                              
            b                              

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