安装 Steam
登录
|
语言
繁體中文(繁体中文)
日本語(日语)
한국어(韩语)
ไทย(泰语)
български(保加利亚语)
Čeština(捷克语)
Dansk(丹麦语)
Deutsch(德语)
English(英语)
Español-España(西班牙语 - 西班牙)
Español - Latinoamérica(西班牙语 - 拉丁美洲)
Ελληνικά(希腊语)
Français(法语)
Italiano(意大利语)
Bahasa Indonesia(印度尼西亚语)
Magyar(匈牙利语)
Nederlands(荷兰语)
Norsk(挪威语)
Polski(波兰语)
Português(葡萄牙语 - 葡萄牙)
Português-Brasil(葡萄牙语 - 巴西)
Română(罗马尼亚语)
Русский(俄语)
Suomi(芬兰语)
Svenska(瑞典语)
Türkçe(土耳其语)
Tiếng Việt(越南语)
Українська(乌克兰语)
报告翻译问题














Thank you
You correctly identified this as the critical number. The first parameter is "OpCodes.Ldc_R4", which tells the computer to push the operand (1.33f) onto the stack. On the next line, you can see it yields "OpCodes.Div", which is the division operator. To see this as a percent, assume you have 1 already on the stack, as the amount of energy in the battery. So the VM does 1/1.33, which leaves 0.75 on the stack. (Obviously, that number gets scaled by whatever number the battery returned).
Bottom line: If you want to have it only take 25% of your stored energy, change that 1.33f to 4.0f.
I saw 1.33f in the code.
What does this mean?
That's better