What does -c option do in GCC [closed]

Answer 1

-c will instruct gcc to only compile the source file to an .o (object) file but does not invoke the linker.

With a project containing many .c files one will typically compile first all .c files to .o files and then link everything together with the libraries.

-c Compile or assemble the source files, but do not link. The linking stage simply is not >done. The ultimate output is in the form of an object file for each source file.

      By default, the object file name for a source file is made by replacing the suffix >.c, .i, .s, etc., with .o.

      Unrecognized input files, not requiring compilation or assembly, are ignored.

Answer 2

-c

Compile or assemble the source files, but do not link. The linking stage simply is not done. The ultimate output is in the form of an object file for each source file.

By default, the object file name for a source file is made by replacing the suffix .c, .i, .s, etc., with .o. Unrecognized input files, not requiring compilation or assembly, are ignored.

-o file

Place output in file file. This applies regardless to whatever sort of output is being produced, whether it be an executable file, an object file, an assembler file or preprocessed C code. If -o is not specified, the default is to put an executable file in a.out, the object file for source.suffix in source.o, its assembler file in source.s, a precompiled header file in source.suffix.gch, and all preprocessed C source on standard output.

More can be found in GCC Manual Page

Answer 3

From man gcc:

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-c Compile or assemble the source files, but do not link. The linking stage simply is not done. The ultimate output is in the form of an object file for each source file.

By default, the object file name for a source file is made by replacing the suffix .c, .i, .s, etc., with .o.

Unrecognized input files, not requiring compilation or assembly, are ignored.

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