C中的指针算法

Question

我有以下代码.也许我没有理解指针算法以及我应该有但是为什么int_pointer增加4而不是1？使用char_pointer,为什么它不会增加4而不是1？

 #include <stdio.h>

 int main() {
    int i;

    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    int int_array[5] = {1, 2, 3, 4, 5};

    char *char_pointer;
    int *int_pointer;

    char_pointer = int_array; // The char_pointer and int_pointer now
    int_pointer = char_array; // point to incompatible data types.

    for(i=0; i < 5; i++) { // Iterate through the int array with the int_pointer.
        printf("[integer pointer] points to %p, which contains the char '%c'\n",
            int_pointer, *int_pointer);
        int_pointer = int_pointer + 1;
    }

    for(i=0; i < 5; i++) { // Iterate through the char array with the char_pointer.
        printf("[char pointer] points to %p, which contains the integer %d\n",
            char_pointer, *char_pointer);
        char_pointer = char_pointer + 1;
    }
 }

OUTPUT:

[integer pointer] points to 0xbffff810, which contains the char 'a'
[integer pointer] points to 0xbffff814, which contains the char 'e'
[integer pointer] points to 0xbffff818, which contains the char ' '
[integer pointer] points to 0xbffff81c, which contains the char '
[integer pointer] points to 0xbffff820, which contains the char ' '
[char pointer] points to 0xbffff7f0, which contains the integer 1
[char pointer] points to 0xbffff7f1, which contains the integer 0
[char pointer] points to 0xbffff7f2, which contains the integer 0
[char pointer] points to 0xbffff7f3, which contains the integer 0
[char pointer] points to 0xbffff7f4, which contains the integer 2

Answer 1

这就是指针aritmetics的工作方式:如果将指针递增1,则地址会增加指针类型的大小.因此,在你的机器中,int是4个字节,递增一个int指针会增加4个字节的地址.

Answer 2

未定义的行为

你所拥有的是未定义的行为,首先你违反了严格的别名规则,这通常使得通过不同类型的指针访问对象是非法的,尽管允许通过char*进行访问.我将在这里引用我的答案,其中更广泛地涵盖了这一点:

虽然允许通过char*访问,但严格的别名规则使得通过不同类型的指针访问对象是非法的.允许编译器假设不同类型的指针不指向相同的存储器并相应地进行优化.它还意味着代码调用未定义的行为,并且可以真正做任何事情.

第二个不同的指针可能有不同的对齐要求,因此通过int指针访问char数组可能会违反此要求,因为char数组可能无法正确对齐int.在C99标准草案涵盖了这部分6.3.2.3 指针它说(重点煤矿):

指向对象或不完整类型的指针可以转换为指向不同对象或不完整类型的指针.如果指向的类型的结果指针未正确对齐⁵⁷⁾,则行为未定义.

一个带有正确标志的好编译器在这里应该有很多帮助,使用clang和以下标志-std=c99 -fsanitize=undefined -Wall -Wextra -Wconversion -pedantic我看到以下警告(见它现场):

warning: incompatible pointer types assigning to 'char *' from 'int [5]' [-Wincompatible-pointer-types]
char_pointer = int_array; // The char_pointer and int_pointer now
             ^ ~~~~~~~~~

warning: incompatible pointer types assigning to 'int *' from 'char [5]' [-Wincompatible-pointer-types]
int_pointer = char_array; // point to incompatible data types.
            ^ ~~~~~~~~~~

并在运行时我看到以下错误:

runtime error: load of misaligned address 0x7fff48833df3 for type 'int', which requires 4 byte alignment
0x7fff48833df3: note: pointer points here
00  e0 3e 83 61 62 63 64 65  00 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  6d 47 60 5a 1d 7f 00
             ^ 

指针算术

因此,指针算法基于所指向类型的大小.如果数组访问基本上是指针算术的语法糖,那么如何才能正常工作？您可以在此处阅读更详细的说明和相关讨论.