可以有人解释这个短段的C ++代码，我不能做头或尾的它 [英] can someone explain this short segment of C++ code, I can't make heads or tails of it

问题描述

#include <opcodes.h>
const char *getOpcodeName(
    uint8_t op
)
{
    #define OPCODE(x, y) if((0x##y)==op) return "OP_" #x;
        OPCODES
    #undef OPCODE
    return "OP_UNKNOWN";
}

链接到此处的代码： https://github.com/znort987/blockparser/blob/master/opcodes.cpp

这是指向包含的 opcodes.h的链接

我知道这只是一个奇怪的格式化函数，但是，我想知道什么是 * 在函数名的开头表示。我假设它与指针有关系。

I understand this is just a strangely formatted function, however, I am wondering what exactly the * at the beginning of the function name means. I assume it has something to do with pointers?

此外， #undef 和 #define 语句有效吗？在任一个之后没有分号，其中一个似乎被定义为一行函数。 （0x ## y）是什么意思？ 返回OP_#x 是什么意思？我从来没有遇到过这样的语法。

Also, how are the #undef and #define statements valid? There's no semicolon after either one, and one of them seems to be defined as a one-line function. What does (0x##y) mean? What does return "OP_" #x mean? I haven't ever come across syntax like this before.

我想更加认真地进入C ++，但是看代码来说明是什么发生了很困难。我如何最有效地学习语法和规则？

I want to get into C++ more seriously, but it's very hard when looking at code to tell what exactly is going on. How can I learn the syntax and rules most effectively?

推荐答案

通过 C ++预处理器，例如使用 g ++ -Wall -C -E opcodes.cpp> opcodes.i 然后查看生成的 opcodes.i

Run your code thru a C++ preprocessor, e.g. using g++ -Wall -C -E opcodes.cpp > opcodes.i then look inside the generated opcodes.i

#define 不是语句，而是预处理器指令

宏 OPCODES 被扩展为一些大块，主要包含 OPCODE NOP，61），其将扩展为类似于

The macro OPCODES gets expanded to some big chunk, notably containing OPCODE( NOP, 61) which would get expanded to something like

if ((0x61)==op) return "OP_" "NOP";

两个字符串字符串连接成一个OP_NOP这里。

The two string literals are concatenated into one, "OP_NOP" here.

GCC有一个很好的文档 cpp预处理器。了解字符串（单个＃像 OPCODE 宏的结尾 #x; ）和 concatenation （双重 ## 如<$ c OPCODE 宏的$ c>（0x ## y））。

GCC has a good documentation on its cpp preprocessor. Read about stringification (with the single # like the ending #x; of the OPCODE macro) and about concatenation (with a double ## like (0x##y) of the OPCODE macro).

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