建模如果我有不具有输入和输出没有像的Modelica - 映射非Modelica的函数来
void foo(void)
{
variable1;
variable2;
for loop
{
if conditions
}
variable2=foobar(); // another function call, foobar() is not modelica function
}
然后我可以在Modelica的它像下面建模非Modelica的功能?
model foo
variable1;
variable2;
algorithm
for loop
{
if conditions
}
variable2 :=foobar(); //foobar here is modelica function
end foo;
谢谢我编辑了我的帖子@Dietmar Winkler – shilu