在没有直接赋值函数的情况下,将一个变量的值复制到另一个变量的最佳方法是什么?
变量是2个字节的定点数,最大值为2147483.647。可用的运算符(为清晰起见重新格式化):
add(<variable>, <constant>)
为给定变量subtract(<variable>, <constant>)
减去multiply(<variable>, <constant>)
乘以divide(<variable>, <constant>)
除以check(<variable1>, <op>, <variable2/constant>)
使用运算符variable1
将<variable2/constant>
的值与op
的值进行比较,并返回一个布尔值。 op
可以是=
,<
,<=
,>=
或>
set(<variable>, <constant>)
设置变量的值。while
和if
(但不是else
)以及基本布尔运算符AND
,OR
,NOT
和NOR
。我们的第一次尝试就是:
while = {
count = var1 //loop var1 times
change_variable = { which = var2 value = 1 } //increment by one
}
我的下一次尝试是:
set_variable = { which = new value = 1 }
while = {
limit = { NOT = { check_variable = { which = old value = new} } }
if = {
limit = {
check_variable = { which = old value > new }
check_variable = { which = stage value = 0 }
}
multiply_variable = { which = new value = 10 }
}
if = {
limit = {
check_variable = { which = stage value = 0 }
check_variable = { which = old value < new }
}
set_variable = { which = stage value = 1 }
}
if = {
limit = {
check_variable = { which = stage value = 1 }
check_variable = { which = old value < new }
}
subtract_variable = { which = new value = 1 }
}
}
例如,从1开始,乘以10直到大于目标,然后减去1直到达到目标。
但我确信有更好的方法可以做到这一点。
上下文
最近在游戏中使用的脚本语言补丁&#34; Stellaris&#34;破坏了变量赋值函数。上面的set
应为set(<variable>, <variable/constant>
)。这已经破坏了许多现有的代码,并且不知道它会在多长时间内被破坏。
答案 0 :(得分:3)
因为2147483.647是(2 - 1的幂)/ 1000我可以使用2的连续幂来一次接近1位解决方案:(使用等效操作用C语言编写)
result = 0;
result += 1073741.824;
if( var < result )
result -= 1073741.824;
result += 536870.912;
if( var < result)
result -= 536870.912;
result += 268435.456;
if( var < result)
result -= 268435.456;
result += 134217.728;
if( var < result)
result -= 134217.728;
result += 67108.864;
if( var < result)
result -= 67108.864;
result += 33554.432;
if( var < result)
result -= 33554.432;
result += 16777.216;
if( var < result)
result -= 16777.216;
result += 8388.608;
if( var < result)
result -= 8388.608;
result += 4194.304;
if( var < result)
result -= 4194.304;
result += 2097.152;
if( var < result)
result -= 2097.152;
result += 1048.576;
if( var < result)
result -= 1048.576;
result += 524.288;
if( var < result)
result -= 524.288;
result += 262.144;
if( var < result)
result -= 262.144;
result += 131.072;
if( var < result)
result -= 131.072;
result += 65.536;
if( var < result)
result -= 65.536;
result += 32.768;
if( var < result)
result -= 32.768;
result += 16.384;
if( var < result)
result -= 16.384;
result += 8.192;
if( var < result)
result -= 8.192;
result += 4.096;
if( var < result)
result -= 4.096;
result += 2.048;
if( var < result)
result -= 2.048;
result += 1.024;
if( var < result)
result -= 1.024;
result += 0.512;
if( var < result)
result -= 0.512;
result += 0.256;
if( var < result)
result -= 0.256;
result += 0.128;
if( var < result)
result -= 0.128
result += 0.064;
if( var < result)
result -= 0.064;
result += 0.032;
if( var < result)
result -= 0.032;
result += 0.016;
if( var < result)
result -= 0.016;
result += 0.008;
if( var < result)
result -= 0.008;
result += 0.004;
if( var < result)
result -= 0.004;
result += 0.002;
if( var < result)
result -= 0.002;
result += 0.001;
if( var < result)
result -= 0.001;
只需要进行32次比较(如果精度较低或知道上限,则可以减少比较)。