有什么比使用静态整数更好的信息存储方式？ C ++
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我通过将玩家的工作设置为数字来跟踪他的“工作”，如果改变工作将其递增一个，并通过数字是偶数还是奇数来确定他当前所从事的工作。 （现在只有两项工作）。但是，我知道这样做的更好方法，很快我将需要执行第三和第四项工作，因此我无法继续使用偶/奇检查。
这是我的代码供参考：（请注意，我只包含相关代码）
GameModeState.cpp
// If changeJob\'s parameter number is 1, it increments the job. If number is 2, it only returns the current job
int GameModeState::changeJob(int number)
{
// Default job is even (landman)
static int job = 1;
if (number == 1)
{
job = (job+1);
return job;
}
else
{
return job;
}
}
int GameModeState::getJob()
{
int currentJob = (changeJob(2));
return currentJob;
}
// If the player opens the \"stat sheet\", it changes their job
void GameModeState::_statSheet(const String& message, const Awesomium::JSValue& input, Awesomium::JSValue& output)
{
changeJob(1);
}
class GameModeState : public GameState::State
{
public:
/// Changes the player\'s job if number is 1, or returns current job if number is 2
static int changeJob(int number);
/// Returns the current job number by calling changeJob appropriately
static int getJob();
private:
// Opening the player sheet will change the player\'s job
void _statSheet(const String& message, const Awesomium::JSValue& input, Awesomium::JSValue& output);
};
#include \"GameModeState.h\"
#include <EnergyGraphics/ZoneParser.h>
void ZoneMovementState::_changeZone(const String& message, const Awesomium::JSValue& input, Awesomium::JSValue& output)
{
// If the number from getJob is even, the player is currently a geologist
if (GameModeState::getJob()%2 == 0)
{
ZoneParser::getSingleton().load(\"../media/zones/geology_zone.xml\", false);
}
else //otherwise they are a landman
{
ZoneParser::getSingleton().load(\"../media/zones/landman_zone.xml\", false);
}
transitionHandler->go();
}
没有找到相关结果
已邀请:
3 个回复
柑恫祟
enum PlayerJob JOB_NONE = 0, JOB_GEOLOGIST, JOB_LANDMAN, ... NUM_JOBS // this element is optional but can be useful for range checking. }; ... PlayerJob job = JOB_NONE; ... switch(job) { case JOB_NONE: break; case JOB_GEOLOGIST: ... break; ... default: error(\"Unhandled palyer job: %d\", job); break; }惭法搽
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