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copied!<p><strong>Short answer:</strong> This function converts opcodes to string.</p> <p>The <code>*</code> IS RELATED to do with pointers! in fact, this function returns a <code>const char *</code> type. This is a pointer for a char buffer (in this case) used to point to a <a href="http://en.wikipedia.org/wiki/Null-terminated_string" rel="nofollow">C-String</a>. Every C-String is actually a buffer with "readable characteres" (alphanumerical, some accents, basic symbols + some stuff) followed by a byte of value 0 (or <code>'\0'</code>) to indicate the end of the string!</p> <p>This function converts opcodes (assembly instructions) to a readable string. So The intention of the programmer was to transform:</p> <ul> <li>0x01 -> OP_CODE1</li> <li>0x02 -> OP_CODE2</li> <li>0x03 -> OP_CODE3</li> </ul> <p>The expanded version of the code is something like this:</p> <pre><code>const char *getOpcodeName( uint8_t op ) { if((0x01)==op) return "OP_X"; if((0x02)==op) return "OP_Y"; if((0x03)==op) return "OP_Z"; ... if((0x??)==op) return "OP_?"; return "OP_UNKNOWN"; } </code></pre> <p>Insted of write houndreds of IF's ... the programmer decided to create the macro</p> <pre><code>#define OPCODE(x, y) if((0x##y)==op) return "OP_" #x; </code></pre> <p>So it's possible to write the OPCODES table easily like this:</p> <pre><code>#define OPCODES \ OPCODE( 01, "X" ) \ OPCODE( 02, "Y" ) \ OPCODE( 03, "Z" ) \ ... OPCODE( ??, "?" ) </code></pre> <p>Opcodes are codes that indentify the processor instruction (assembly instructions). Example of real instructions (for some <a href="http://sparksandflames.com/files/x86InstructionChart.html" rel="nofollow">Intel processor</a>) are:</p> <pre><code> inc eax ; opcode = 0x40 pusha ; opcode = 0x60 nop ; opcode = 0x90 </code></pre> <p>So your table could be:</p> <pre><code>#define OPCODES \ OPCODE( 40, "INCEAX" ) \ OPCODE( 60, "PUSHA" ) \ OPCODE( 90, "NOP" ) </code></pre>

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