C++指针算术怪异

Question

我发现了我的错误(几个小时后)并将其隔离在以下程序中。问题在于使用指向结构的指针时计算 pst2 变量值的方式。使用指向 char 的指针时，一切正常。这是为什么？
(使用 gcc/g++ 版本:(Debian 4.4.5-8)4.4.5)
(对于那些想知道的人:我正在访问一个包含数据分组的文件缓冲区以常规偏移量。)

#include <iostream>
#include "testpa.h"

#pragma pack(push)
#pragma pack(1)
//---------------------------
struct st_one
{
    int i;
    char c;
};
//---------------------------
struct st_two
{
    long l;
    int i;
};
#pragma pack(pop)

//===========================
int main()
{
    int n=1024, np1=sizeof(st_one); //, np2=sizeof(st_two);
    st_one *pst1, *pst1a;
    st_two *pst2, *pst2a;
    char *pc1, *pc2, *pc1a, *pc2a, *pb;

    pb = new char[n];

    pst1 = (st_one*)(pb);
    pst2 = (st_two*)(pst1 + np1); //using pst1
    pc1 = (char*)(pb);
    pc2 = (char*)(pc1 + np1); //using pc1

    pst1a = (st_one*)(pb);
    pst2a = (st_two*)(pb + np1); //using pb
    pc1a = (char*)(pb);
    pc2a = (char*)(pb + np1); //using pb

    std::cout << "\npb = " << (long)pb;
    std::cout << "\n-----";
    std::cout << "\npst1 = " << (long)pst1 << "\tpst2 = " << (long)pst2;
    std::cout << "\npc1  = " << (long)pc1 << "\tpc2  = " << (long)pc2;
    std::cout << "\n-----";
    std::cout << "\npst1a = " << (long)pst1a << "\tpst2a = " << (long)pst2a;
    std::cout << "\npc1a  = " << (long)pc1a << "\tpc2a  = " << (long)pc2a;
    std::cout << "\n-----\n";

    return 0;
}

输出:

pb = 19546128

pst1 = 19546128         pst2 = 19546153  <--- WRONG!
pc1  = 19546128         pc2  = 19546133

pst1a = 19546128        pst2a = 19546133
pc1a  = 19546128        pc2a  = 19546133

Answer 1

我觉得不错。线路:

 (pst1 + np1)

将 st_one 的 np1 实例添加到 pst1 指向的地方，这意味着 pst1 的值增加by np1 * sizeof (st_one) 字节，即 25 (sizeof = 5)，对应于您输出的值。而不是上面的，我想你想要的是:

 (pst1 + 1)

pc1 值起作用是因为它是一个 char 指针，所以行:

(pc1 + np1)

将 np1 * sizeof (char) 字节添加到 pc1，即 5 个字节。

递增指针使指针指向内存中的下一个元素，而不是下一个字节。