编译器错误“在令牌之前缺少二元运算符"是什么意思?意思是? [英] What does the compiler error "missing binary operator before token" mean?

问题描述

我最近在尝试使用 gcc 进行编译时遇到以下错误:

I recently got the following error when trying to compile with gcc:

错误:在标记("之前缺少二元运算符

error: missing binary operator before token "("

Web 和 SO 搜索提出了此错误的几个具体示例，并更改了特定代码以修复它们.但是我没有找到什么条件导致这个错误的一般描述.

Web and SO searches came up with several specific examples of this error, with specific code changes to fix them. But I found no general description of what condition causes this error to be issued.

gcc 何时以及为什么会发出此错误?

When and why does gcc emit this error?

推荐答案

这不是编译器错误，而是预处理器错误.当预处理器在尝试评估 #if 或 #elif 指令中的表达式时遇到无效语法时，就会发生这种情况.

This is not a compiler error, it is a preprocessor error. It occurs when the preprocessor encounters invalid syntax while trying to evaluate an expression in a #if or #elif directive.

一个常见的原因是 #if 指令中的 sizeof 操作符:

One common cause is the sizeof operator in an #if directive:

例如:

  #define NBITS (sizeof(TYPE)*8)
  //later
  #if (NBITS>16)    //ERROR

这是一个错误，因为 sizeof 是由编译器计算的，而不是预处理器.

This is an error because sizeof is evaluated by the compiler, not the preprocesor.

类型转换也不是有效的预处理器语法:

Type casts are also not valid preprocessor syntax:

  #define ALLBITS ((unsigned int) -1)
  //later
  #if (ALLBITS>0xFFFF)    //ERROR

有效表达式的规则在这里.

另请注意，#if 会将未定义的宏计算为 0，除非它看起来需要参数，在这种情况下，您 也会出现这个错误:

Note also that #if will evaluate an undefined macro as 0, unless it looks like it takes arguments, in which case you also get this error:

所以如果 THIS 未定义:

#if THIS == 0  //valid, true

#if THIS > 0 //valid, false

#if THIS() == 0  //invalid. ERROR

#if 语句中的拼写错误 也可能导致此消息.

Typos in your #if statement can also cause this message.

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