embedded

This query is regarding allocation of memory using malloc . Generally what we say is malloc allocates memory from heap. Now say I have a plain embedded system(No operating system), I have normal program loaded where I do malloc in my program. In this case where is the memory allocated from ? malloc() is a function that is usually implemented by the runtime-library. You are right, if you are running on top of an operating system, then malloc will sometimes (but not every time) trigger a system-call that makes the OS map some memory into your program's address space. If your program runs without

We are considering porting an application from a dedicated digital signal processing chip to run on generic x86 hardware. The application does a lot of Fourier transforms, and from brief research, it appears that FFTs are fairly well suited to computation on a GPU rather than a CPU. For example, this page has some benchmarks with a Core 2 Quad and a GF 8800 GTX that show a 10-fold decrease in calculation time when using the GPU: http://www.cv.nrao.edu/~pdemores/gpu/ However, in our product, size constraints restrict us to small form factors such as PC104 or Mini-ITX, and thus to rather limited

I have a PID controller running on a robot that is designed to make the robot steer onto a compass heading. The PID correction is recalculated/applied at a rate of 20Hz. Although the PID controller works well in PD mode (IE, with the integral term zero'd out) even the slightest amount of integral will force the output unstable in such a way that the steering actuator is pushed to either the left or right extreme. Code: private static void DoPID(object o) { // Bring the LED up to signify frame start BoardLED.Write(true); // Get IMU heading float currentHeading = (float)RazorIMU.Yaw; // We just

I was reading http://embeddedgurus.com/embedded-bridge/2010/03/different-bit-types-in-different-registers/ , which said: With read/write bits, firmware sets and clears bits when needed. It typically first reads the register, modifies the desired bit, then writes the modified value back out and I have run into that consrtuct while maintaining some production code coded by old salt embedded guys here. I don't understand why this is necessary. When I want to set/clear a bit, I always just or/nand with a bitmask. To my mind, this solves any threadsafe problems, since I assume setting (either by

I am debugging this piece of software for an STM32 embedded system. In one of the functions my programs keeps hitting some sort of breakpoint: SIGTRAP, Trace/breakpoint trap However, in GDB, when I do info breakpoints I get No breakpoints or watchpoints . The breakpoint actually corresponds to a breakpoint I had set quite some time ago, in another version of the executable. When I set that breakpoint, GDB told me automatically using a hardware breakpoint on read-only memory (or a similar message). I think the hardware breakpoint remains on my chip, despite having loaded a new version of the

I started using MPLAB recently, but for someone that works with Eclipse and VS the IDE it's very limited. Do you know any free IDE or how to configure Ecplise or Netbeans to PIC development? Thanks all The underlying toolchain (compiler/linker etc.) can be used from any environment including Eclipse and Visual Studio, though Eclipse is probably the more flexible in this respect. MPLAB has a feature to export a project as a makefile that can be used with GNU make, although you may rather generate your own makefile, or use the project management provided by Eclipse. In Visual Studio, create a

There are a lot of questions about memory allocation on this site, but I couldn't find one that specifically addresses my concern. This question seems closest, and it led me to this article , so... I compared the behavior of the three demo programs it contains on a (virtual) desktop x86 Linux system and an ARM-based system. My findings are detailed here , but the quick summary is: on my desktop system, the demo3 program from the article seems to show that malloc() always lies about the amount of memory allocated—even with swap disabled. For example, it cheerfully 'allocates' 3 GB of RAM, and