长不等于另一个长C++

Question

当0 == 0函数返回 long 和 regularlong 的 sizeof（） 都说 8 时，它永远不会进入 if 语句，我不知道这里还有什么问题。

声明：

long                        RemoteStep; // Next packet number to-be-processed
long GetLong(BYTE * Message, const SHORT Offset)
{   // Get a long from a char *
    return *(long*)&(Message[Offset]);
}

调试代码：

printf("id = %d  remotestep = %d n", GetLong(Packet->Message, 2), RemoteStep);
printf("id = %d  remotestep = %d n", GetLong(Packet->Message, 2), RemoteStep);
printf("id = %d  remotestep = %d n", GetLong(Packet->Message, 2), RemoteStep);
printf("equals = %d n", GetLong(Packet->Message, 2) == RemoteStep);
printf("sizeof = %d - %dn", sizeof(GetLong(Packet->Message, 2)), sizeof(RemoteStep));
        // Process if expected
        if (GetLong(Packet->Message, 2) == RemoteStep)
        {
            printf("in.............n");
            ...
        }

输出信息：

id = 0  remotestep = 0
id = 0  remotestep = 0
id = 0  remotestep = 0
equals = 0
sizeof = 8 - 8
id = 1  remotestep = 0
id = 1  remotestep = 0
id = 1  remotestep = 0
equals = 0
sizeof = 8 - 8

我在compat-gcc-34-c++又名 g++34 下编译了它，我不能使用较新的 g++ 编译器，因为它们给出了太多的警告甚至错误。

与-Wall -Wextra

declares.h:1880: warning: int format, different type arg (arg 2)
declares.h:1880: warning: int format, different type arg (arg 3)
declares.h:1880: warning: unknown conversion type character 0x20 in format
declares.h:1880: warning: unknown conversion type character 0x20 in format
declares.h:1880: warning: too many arguments for format
declares.h:1881: warning: unknown conversion type character 0x20 in format
declares.h:1881: warning: unknown conversion type character 0x20 in format
declares.h:1881: warning: too many arguments for format
declares.h:1882: warning: unknown conversion type character 0x20 in format
declares.h:1882: warning: unknown conversion type character 0x20 in format
declares.h:1882: warning: too many arguments for format
declares.h:1884: warning: int format, different type arg (arg 2)
declares.h:1884: warning: int format, different type arg (arg 3)

1880行是其中之一

printf("id = %l  remotestep = %l n", GetLong(Packet->Message, 2), RemoteStep);

Answer 1

long 类型不应与 %d 一起打印，这在规范中明确未定义。例如

每个转换规范都由"%"字符引入...
...
如果转换规范与上述形式之一不匹配，则行为未定义。如果任何参数不是相应转换规范的正确类型，则行为未定义。 http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html

（文档适用于printf以及fprintf）

---

因此，您不能依赖 的输出，

printf("id = %d  remotestep = %d n", GetLong(Packet->Message, 2), RemoteStep);

以确定

printf("equals = %d n", GetLong(Packet->Message, 2) == RemoteStep);
//output: equals = 0

实际上是不正确的，您需要先修复调试语句：

printf("id = %ld  remotestep = %ld n", GetLong(Packet->Message, 2), RemoteStep);

---

在您的情况下，最好使用 memcpy 而不是指针转换，

long GetLong(BYTE * Message, const SHORT Offset)
{   
    long result;
    std::memcpy(&result, &(Message[Offset]), sizeof(long)); 
    return result;
}

因为long*在您的平台上可以有比char*更严格的对齐要求.

Answer 2

您可以将语句更改为使用 std：：cout 而不是 printf，因为您使用 C++ 而不是 C 标记了问题。使用 cout，您将不会在运行时获得 %d 或其他 % 说明符的这些令人惊讶的效果。

printf("id = %d remotestep = %d n", GetLong(Packet->Message, 2), RemoteStep);

大概可以替换为

std::cout << "id = " << GetLong( Packet->Message, 2 ) << " remotestep = " << RemoteStep << " n";