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This morning , I was trying to compile a C++ program using the following command on the command line

g++ -o foo.cpp foo

I spend about an hour trying t to resolve why I kept getting undefined reference & multiple definition error till I realised that my command to compile was wrong and it should be

g++ foo.cpp -o foo

I am just wondering what is the difference in process between g++ -o foo.cpp foo and g++ foo.cpp -o foo .

Why does placing the -o infront cause compilation error

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2 Answers 2

Because -o specifies the outfile. The first means that the outfile would be 'foo.cpp' whereas the second means that your outfile will be 'foo'.

The syntax of g++ is:

gcc [-c|-S|-E] [-std=standard] [-g] [-pg] [-Olevel] [-Wwarn...] [-pedantic] [-Idir...] [-Ldir...] [-Dmacro[=defn]...] [-Umacro] [-foption...] [-mmachine-option...] [-o outfile] [@file] infile...

Relevant man page: http://linux.die.net/man/1/g++

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The switch -o takes an argument. In your case you were telling g++ to output the resulting compilation to the file foo.cpp. That's not what you want.

$ g++ -o foo.cpp foo

You want to output the resulting compilation of the source file (foo.cpp) to the binary file foo.

$ g++ foo.cpp -o foo

excerpt from g++ man page

-o <file>                Place the output into <file>
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