Question

In the following example, the program should print "foo called":

// foo.c
#include <stdio.h>

__attribute__((constructor)) void foo()
{
    printf("foo called\n");
}

// main.c
int main()
{
    return 0;
}

If the program is compiled like this, it works:

gcc -o test main.c foo.c

However, if foo.c is compiled into a static library, the program prints nothing.

gcc -c main.c
gcc -c foo.c
as rcs foo.a foo.o
gcc -o test foo.a main.o

Why does this happen?

Was it helpful?

Solution

The linker does not include the code in foo.a in the final program because nothing in main.o references it. If main.c is rewritten as follows, the program will work:

//main.c

void foo();

int main()
{
    void (*f)() = foo;
    return 0;
}

Also, when compiling with a static library, the order of the arguments to gcc (or the linker) is significant: the library must come after the objects that reference it.

gcc -o test main.o foo.a

OTHER TIPS

As it was stated, unreferenced symbols from archive does not make it to the output binary, because linker discards them by default.

To override this behaviour when linking with static library, --whole-archive/--no-whole-archive options for the linker may be used, like this:

gcc -c main.c
gcc -c foo.c
ar rcs foo.a foo.o
gcc -o test -Wl,--whole-archive foo.a -Wl,--no-whole-archive main.o

This may lead to bloated binary, because all symbols from foo.a will be included by the linker to the output, but sometimes it is justified.

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top