Eclipse DEBUG error:can't find source file at "cyu3device.c"

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Anonymous
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Hello guys,

   

I am very new in using Eclipse and trying to debug for the first time .

   

I have tried googling it and found many solutions but didnt understand anything.

   

I know its a very small problem. Can you please tell me what to do+

   

I am using cypress sdk 1.2.3 . I have my cypress sdk located in C directory and my workspace is in D directory.

   

The project which I am working on is made on a older version.

   

If you need any more information please dont hesitate to ask

   

Kind Regards,

   

Pruthvi

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9 Replies
Anonymous
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This is extensively documented in section 12.2.2.3 of the FX3 Programmer's Manual. Read that and see if you can successfully debug project.

   

-- Manu

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Anonymous
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Hello

   

Thanks for your reply

   

yes I have made all the settings suggested by programmers manual

   

after I start executing step by step that is when I am debugging when it is at the step

   

   

status =

       

   

CyU3PDeviceInit (&clkCfg);

   

it gives the error

   
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Anonymous
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Hello Manu,

   

sorry I mean to say resume executing as similiar to step "12.2.2.3 page no 164".

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Anonymous
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 Raj

   

The source of APIs are not available. So while step-by-step debuggin when you try to enter an API, it throws that error that the source is not available. Ignore that and  Click on 'resume' or 'step-over' button.

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Anonymous
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 CyU3PDeviceInit() is implemented in a file called cyu3device.c in the SDK. But we don't release the source code and so you can't jump into the function.

   

So, you can use 'Step Into' only for your functions or the ones already there in the firmware. For SDK functions, use 'Step Over'.

   

-- Manu

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Anonymous
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Thanks for the help NIKL and MANU

   

I am already working on it.

   

Can some one explain me what does this way of gpio settings mean

   

   

io_cfg.

       

   

gpioSimpleEn[1] = (1 << (FPGA_DONE-32)) | (1 << (FPGA_INIT_N-32))

   

| (1 << (DQ16-32))| (1 << (DQ17-32))| (1 << (DQ18-32))| (1 << (DQ19-32))| (1 << (DQ20-32))

   

| (1 << (DQ21-32))| (1 << (DQ22-32))| (1 << (DQ23-32)) | (1 << (DQ24-32))| (1 << (DQ25-32))

   

| (1 << (DQ26-32))| (1 << (DQ27-32));

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Anonymous
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io_cfg.gpioSimpleEn[1] = 1<<SOME_PIN means that SOME_PIN is being used as a Simple GPIO. gpioSimpleEn is a bitmapped variable where a '1' at bit 'x' means that pin GPIO[x] is being used for that purpose (simple GPIO). Since there are >32 pins, gpioSimpleEn is an array of 2 uint32s.

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Anonymous
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Hello Manu

   

Thanks  for the information you made things very clear

   

 I am presently programming a firmware to perform slavefifo sync interface with FPGA.

   

So should I also initialize using the gpiosimpleen the clock signal from fpga, slcs,slrd etc etc...??

   

Kind Regards

   

Pruthvi

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Anonymous
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 No. These signals are control lines that are used by the GPIF hardware. If you initialize them in gpioSimpleEn, they will be used as GPIO and not control lines for GPIF.

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