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questions about objcopy and -O option
- From: ali hagigat <hagigatali at gmail dot com>
- To: binutils at sourceware dot org
- Date: Sun, 6 Feb 2011 09:07:59 +0330
- Subject: questions about objcopy and -O option
When i use: objcopy -O binary....The man pages of objcopy says:
"When objcopy generates a raw binary file, it will
essentially produce a memory dump of the contents of the input object file. All
symbols and relocation information will be discarded. The memory dump will
start at the load address of the lowest section copied into the output file."
By using objcopy i am creating an output file. So what is memory dump?
Does it create any thing in memory?
What is the content of this file? Only translated instructions, or it
contains some extra reserved bytes? Suppose i had a ld script and set
the start of my sections at 0xffffff00, after using objcopy it seems
that all this information is removed. Is that right?
Does compiler add any extra instructions to my code to remember those
addresses? I mean by looking at the output file of objcopy I can see
the address of 0xffffff00?