#include <stdio.h>
// xyz will be emitted with -flto (or if it is static) even when
// the function is unused
__attribute__((__used__))
void xyz() {
printf("Hello World!\n");
}
int main() {
return 0;
}
What do I need this for?
Is there any way I could still reach xyz somehow besides directly calling the function, like some dlsym() like magic?
Attribute used is helpful in situation when you want to force compiler to emit symbol, when normally it may be omitted. As GCC's documentation says (emphasis mine):
This attribute, attached to a function, means that code must be emitted for the function even if it appears that the function is not referenced. This is useful, for example, when the function is referenced only in inline assembly.
For instance, if you have code as follows:
#include <iostream>
static int foo(int a, int b)
{
return a + b;
}
int main()
{
int result = 0;
// some inline assembly that calls foo and updates result
std::cout << result << std::endl;
}
you might notice, that no symbol foo is present with -O flag (optimization level -O1):
g++ -O -pedantic -Wall check.cpp -c
check.cpp:3: warning: ‘int foo(int, int)’ defined but not used
nm check.o | c++filt | grep foo
As a result you cannot reference foo within this (imaginary) inline assembly.
By adding:
__attribute__((__used__))
it turns into:
g++ -O -pedantic -Wall check.cpp -c
nm check.o | c++filt | grep foo
00000000 t foo(int, int)
thus now foo can be referenced within it.
You may also have spotted that gcc's warning is now gone, as you have tell you compiler that you are sure that foo is actually used "behind the scene".
A particular usecase is for interrupt service routines in a static library. For example, a timer overflow interrupt:
void __attribute__((interrupt(TIMERA_VECTOR),used)) timera_isr(void)
This timera_isr is never called by any function in the user code, but it might form an essential part of a library. To ensure it is linked and there isn't a interrupt vector pointing to an empty section the keyword ensures the linker doesn't optimise it out.
If you declare a global variable or function that is unused, gcc will optimized it out (with warning), but if you declared the global variable or the function with '__attribute__((used))', gcc will include it in object file (and linked executable).
https://gcc.gnu.org/legacy-ml/gcc-help/2013-09/msg00108.html