search for: __mips16_extendsfdf2

...le. long int XXX_extendsfdf2(float); The problem for me is that the parameter should be int since this is a soft float routine. Probably nobody ever looks at this argument type in this situation. So I think this should look like: long int XXX_extendsfdf2(int); In my case the routine is called __mips16_extendsfdf2 . When I see the first parameter as float, I think that I need to generate a mips16 helper function. This is because in the abi that parameter would be passed in a floating point argument register but this is not possible in mips16 mode since there are no floating point instructions. For inter...

...t; > The problem for me is that the parameter should be int since this is a > soft float routine. Probably nobody ever looks at this argument type in > this situation. > > So I think this should look like: > long int XXX_extendsfdf2(int); > > In my case the routine is called __mips16_extendsfdf2 . > > When I see the first parameter as float, I think that I need to generate > a mips16 helper function. This is because in the abi that parameter > would be passed in a floating point argument register but this is not > possible in mips16 mode since there are no floating point ins...

Here is a test case: extern double floor(double); extern double floor_(double); double x = 1.5; double y, y_; void foo() { double y = floor(x); double y_ = floor_(x); } If I compile this for Mips16, it calls the proper helper function for floor_ but not for floor, because the signature for floor in callee info is wrong. Args[0] = void RetTy = void

I'm getting some problems because it seems that the compiler is treating "floor" differently from other math library functions like "sin". The Args and RetVal have the parameter and return types marked as void. For mips16, it's important that I be able to know the original signature for floating point functions. In some cases, need to create calls to helper