feuling@astroatc.UUCP (Lindsay Feuling) (01/19/91)
On compiling my code with Microsoft Quick C 2.?, I put in the /FPi87 flag in the make file. I'm not sure whether QuickC actually recognizes this flag. Also, in looking at the assembly code generated, I noticed that it does not generate 8087 instructions. What it actually does is generate code to call some math emulator functions via interrupts, but I want to run this code within an Interrupt Service Routine without goofing around with calling other interrupt functions. In looking at the code with CV, I found something of the following: int 37 ;fild dword ptr [FFfac (2009)] int 3d ;fwait les bx,dword ptr [p_temp] int 3c ;fstp dword ptr es:[bx] int 3d ;fwait If it amounts to actually going into the assembly code and changing the assembly code so that the 8087 is used, then I will. But not if there's something that I have missed. Could anyone give me a clue as to what, if any, compiler flag will actually put in 8087 in-line code (for real!) using QuickC? Thanks L. Feuling Astronautics Tech. Ctr. madison, wi