ColdFire/68K Microcontrollers and ProcessorsのトピックRe: mcf5271

mcf5271

Anand — Thu, 24 May 2007 02:35:31 GMT

Hi,

Is MCF5271 a byte addressable processor? If I am dealing with 32 bit data, does the program counter increment by 4?

Re: mcf5271

SimonMarsden_de — Thu, 24 May 2007 15:57:27 GMT

When reading or writing data, you can transfer byte, 2-byte or 4-byte quanties (ColdFire calls these Byte, Word and Long).

Instruction widths are either 2, 4 or 6 bytes. The PC thus increments by 2, 4, or 6 each time a non-branch instruction is executed. Groups of instructions may be fetched from memory in a burst transfer.

Hope this helps

Simon

Re: mcf5271

Anand — Thu, 24 May 2007 23:26:12 GMT

Hi Simon,

Thanks for your reply. I am dealing with 32 bit data. So everytime I fetch 32 bits of data, the program counter would increment by 4 right??

Re: mcf5271

mccPaul — Fri, 25 May 2007 15:18:21 GMT

Hi Anand

What do you mean by "everytime I fetch 32 bits of data"? Are you are talking about loading a register with data or dereferencing or incrementing a C pointer?

The program counter increments as each intruction is executed by the CPU and as Simon said, the instruction length is variable so the program counter can be incremented by 2, 4 or 6 bytes.

We may be able to be clearer if you could show some source code that you need to have explained.

Cheers,

Paul.

Re: mcf5271

SimonMarsden_de — Fri, 25 May 2007 15:27:42 GMT

Hi Anand

I think you might be confusing two separate concepts - the program instructions (opcodes) in memory telling the processor what to do, and the data that the program operates on.

The Program Counter (PC) points to the next opcode to execute. Each time a new instruction is executed, it will increment by 2, 4 or 6 depending on the width of the instruction.

Separate from that, the processor can also read and write data. It can do this in byte, word or long-word transfers. There are many instructions capable of fetching 32-bit data from memory. Here are just a few:

    move.l   (a0),d0

    move.l   4(a3),d0

    move.l   fred(a2,d1.l*8),d0

    add.l     mydata,d0

All of these fetch 32 bits of data (the width is specified as xxx.L in each instruction). However, the widths of the instruction opcodes vary, and so the PC will increment by a variable amount.

Hope this helps

Simon