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在这个程序中，我声明了一个结构和几个成员，并且我使用了一个包含两个元素的数组作为这个特定结构的实例。我在这个结构中使用了一个 5 元素数组。一个特殊的事情是，在主函数中，当我尝试获取 5 元素数组中不存在的第 6 个元素的地址时，我得到了下一个结构的第一个成员整数的地址。为什么当我尝试访问不存在的数组元素时没有显示错误，为什么两个成员具有相同的地址？

#include<stdio.h>
 struct {int num1,num2; char s1; int *ptr; int abc[5]; }a[2];

void main(){

      int start, last; 

      start = &a[1].num1;

      last = &a[0].num1;

      printf("\nSize of the Structure : %d Bytes", start-last);

      printf("\naddress of num1 in structure a[0] is %d", &a[0].num1);

      printf("\naddress of num2 in structure  a[0] is %d", &a[0].num2);

      printf("\naddress of char in structure a[0] is %d", &a[0].s1);

      printf("\naddress of ptr in structure a[0] is %d", &a[0].ptr);

      printf("\naddress of I element in array abc[5] in structure a[0] is %d", &a[0].abc[0]);

      printf("\naddress of II element in array abc[5] in structure a[0] is %d", &a[0].abc[1]);

      printf("\naddress of III element in array abc[5] in structure a[0] is %d", &a[0].abc[2]);

      printf("\naddress of IV element in array abc[5] in structure a[0] is %d", &a[0].abc[3]);

      printf("\naddress of V element in array abc[5] in structure a[0] is %d", &a[0].abc[4]);

      **printf("\naddress of VI element in the 5 element array abc[5] in structure a[0] is %d", &a[0].abc[5]);

      printf("\naddress of num1 in structure a[1] is %d", &a[1].num1);**

      printf("\n");


  }

kevin@kevin-desktop:~/Documents/programs$ gcc --version
gcc (Ubuntu 4.4.3-4ubuntu5.1) 4.4.3
Copyright (C) 2009 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

kevin@kevin-desktop:~/Documents/programs$ gcc structure_size.c -o structure_size
structure_size.c: In function ‘main’:
structure_size.c:15: warning: assignment makes integer from pointer without a cast
structure_size.c:17: warning: assignment makes integer from pointer without a cast
structure_size.c:21: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:23: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:25: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘char *’
structure_size.c:27: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
structure_size.c:29: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:31: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:33: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:35: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:37: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:39: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
structure_size.c:41: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
kevin@kevin-desktop:~/Documents/programs$ ./structure_size

Size of the Structure : 36 Bytes
address of num1 in structure a[0] is 134520896
address of num2 in structure  a[0] is 134520900
address of char in structure a[0] is 134520904
address of ptr in structure a[0] is 134520908
address of I element in array abc[5] in structure a[0] is 134520912
address of II element in array abc[5] in structure a[0] is 134520916
address of III element in array abc[5] in structure a[0] is 134520920
address of IV element in array abc[5] in structure a[0] is 134520924
address of V element in array abc[5] in structure a[0] is 134520928
**address of VI element in the 5 element array abc[5] in structure a[0] is 134520932
address of num1 in structure a[1] is 134520932**
kevin@kevin-desktop:~/Documents/programs$
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3 回答 3

2

这就是 C 的工作方式。我认为这个简化的代码更容易显示这种现象：

char string[2][5];              // allocates a block of 10 bytes

string == string[0];            // true
string == &string[0][0];        // true
&string[0][5] == &string[1][0]; // true
于 2013-11-06T08:31:53.690 回答
1

因为您所做的是未定义的行为，编译器当然不需要检测和报告它。

这是（当然）因为这样做通常会非常困难，并且还因为您可以“动态地”做同样的事情，这使得在编译时或多或少不可能检测到。

请记住，C 没有对数组边界的运行时检查。您可以索引任何您想要的内容，并且在大多数情况下，编译器会发出代码并尝试进行访问。它可能会被操作系统阻止，但它也可能“只是这样做”并对您的生活造成严重破坏。

正如@Adriano 所说，欢迎来到 C！:)

PS 指针应该用 打印%p，它们不是ints。

于 2013-11-06T08:16:51.863 回答
0

您的 abc 数组大小为 5，表示您有 a[0]、a[1]、a[2]、a[3]、a[4]。

你完成了 5 个元素。但是你写了一个[5]。不属于那个结构。

a[2] 隐含为 (a+2) 其中“a”是地址，a = 134520896，

所以 a + 2 是 134520904

数组中有 5 个元素，因此数组将获得 20 个字节。即从 134520896 到 134520915。

134520916 是数组 a 的下一个变量的地址，它不是数组 a 的内存

于 2013-11-06T10:11:58.743 回答