宏if语句返回错误：运算符'&&'没有正确的操作数 [英] macro if statement returns error: operator '&&' has no right operand

问题描述

我正在许多Linux机器上编译代码，在一台特定机器上，我收到以下错误：

I am compiling my code on many linux machines, on a specific machine, I receive the following error:

error: operator '&&' has no right operand

宏代码为：

#if LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49) && KERNEL_PATCH_LEVEL == 11

其中LINUX_VERSION_CODE和KERNEL_VERSION在linux源文件中定义，而KERNEL_PATCH_LEVEL在我的Makefile文件中定义

where LINUX_VERSION_CODE and KERNEL_VERSION are defined in linux sources and KERNEL_PATCH_LEVEL is defined in my Makefile

KERNEL_PATCH_LEVEL :=$(word 1, $(subst ., ,$(word 2, $(subst -, ,$(KERNEL_HEADERS)))))

如果我将代码更改为2个不同的行，例如有效：

If i change the code to 2 different lines, like this, it works:

#if LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49) 
#if KERNEL_PATCH_LEVEL == 11
  ...
#endif //KERNEL_PATCH_LEVEL == 11
#endif 

是否可以将其与#if一起保存？
我使用的是gcc版本4.9.0（Debian 4.9.0-7）

Is it possible to still keep it with one #if ? I use gcc version 4.9.0 (Debian 4.9.0-7)

以下宏无效：

#if (LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49)) && (KERNEL_PATCH_LEVEL == 11)  

#if ((LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49)) && (KERNEL_PATCH_LEVEL == 11))  

#if defined(KERNEL_MAJOR) && defined(KERNEL_MINOR) && defined(KERNEL_MICRO) && defined(KERNEL_PATCH_LEVEL) && defined(KERNEL_VERSION) && 
defined(LINUX_VERSION_CODE) && \
     (LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49)) && (KERNEL_PATCH_LEVEL == 11)

推荐答案

事实证明，错误源是在makefile中定义了 KERNEL_PATCH_LEVEL 但为空。

it turns out that the error source is that KERNEL_PATCH_LEVEL is defined in the makefile but empty.

在这种情况下，两行不是等于t，因为

In that case, the 2 lines aren't equivalent, since

#if LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49) && KERNEL_PATCH_LEVEL == 11

无论第一个测试的结果如何，都对这两个部分进行评估，因此遇到&&时，预处理器偶然发现语法错误== 11 。

evaluates both parts no matter what the the outcome of the first test is, so the preprocessor stumbles on the syntax error when meeting && == 11.

但是如果 LINUX_VERSION_CODE == KERNEL_VERSION（3,12,49）是错误的，具有以下构造：

But if LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49) is false, with this construct:

#if LINUX_VERSION_CODE == KERNEL_VERSION(3,12,49) 
#if KERNEL_PATCH_LEVEL == 11
  ...
#endif //KERNEL_PATCH_LEVEL == 11
#endif 

您没有输入第一个 #if ，因此输入的是内部 #if （这是错误的，因为价值 #if == 11 ）属于预处理器跳过的一个块，这说明没有错误。

you're not entering the first #if so the inner #if (which is wrong since worth #if == 11) belongs to a block which skipped by the preprocessor, which explains that there's no error.

请注意，如果未定义 KERNEL_PATCH_LEVEL ，则 #if 会将其视为 0 ，那不会触发任何错误。

Note that if KERNEL_PATCH_LEVEL is not defined, #if sees that as 0, that wouldn't have triggered any error.

您可以防止定义不明确的 KERNEL_PATCH_LEVEL 与此（参见测试空宏定义，已经添加了更好的答案）

you can protect against ill-defined KERNEL_PATCH_LEVEL with this (seen in Test for empty macro definition, I have added a better answer now)

#if (KERNEL_PATCH_LEVEL + 0) == 0
#undef KERNEL_PATCH_LEVEL
#define KERNEL_PATCH_LEVEL 0
#endif

因此，如果宏定义为空（或为0），请取消定义并将其定义为 0 值。您甚至可以检测到它是否为空（而不是 0 这样）（

so if the macro is defined empty (or is 0), undefine it and define it to a 0 value. You could even detect (see my answer in the link above to understand how it works) if it's empty instead of 0 like this:

#if (0-KERNEL_PATCH_LEVEL-1)==1 && (KERNEL_PATCH_LEVEL+0)!=-2
#error "KERNEL_PATCH_LEVEL defined empty"
#endif

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