I'm working with, solaris-9 and using g++ compiler.
I’m trying to get gprof output for a binary file which includes a shared
library.
So I wrote a library like this,
It contains one .cpp file called comp.cpp.
So I compiled it like this,
g++ -c –fPIC –pg comp.cpp
Then I linked it like this,
ld –G comp.o –o libmitTest.so
Likewise I made a library called libmitTest.so
And I wrote a program with two .cpp files.
Let’s say cat.cpp and zoo.cpp
So I compiled those files like this,
g++ -c –pg cat.cpp
g++ -c –pg zoo.cpp
Then I linked that program like this,
g++ -pg –L. –lmitTest zoo.o cat.o –o y.out
After that I changed the environment variables like this,
export LD_PROFILE=y.out:libmitTest.so
After that using ./y.out I ran the program to generate gmon.out file.
Then I tried to get gprof output using the command.
gprof –b y.out gmon.out
But it only shows details about functions in y.out file only. It won’t show
details about function in that shared library. Instead those library functions
there’s a syntax called <external>. But I want to get details about those
shared library functions.
If I delete that shared library functions in my program, previous <external>
command will disappear from the gprof output.
Could you please tell me what did I do wrong here? I think I did something
wrong while linking my shared library using command,
ld –G comp.o –o libmitTest.so
have I done anything wrong here?
Or please tell me the way to get details about a shared library.
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