why gcc can link objs with multiple same symbols after ar?-优快云博客

本文探讨了GCC编译器在不同情况下如何处理具有相同符号的多个目标文件的链接过程。通过具体的代码示例展示了直接链接与使用归档文件时链接器的行为差异，以及链接顺序对最终结果的影响。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1. why gcc can link objs with multiple same symbols after ar? what's the difference comparing with linking before ar?
2. If multiple same symbols are allowable in linking, then what's the rule to resolve undefined function call(claimed with "extern" before)?

a.c:

extern int f();

int main()

...{

printf("result is %d ",f());

return f();

}

int f()

...{

return 1;

}

b.c:

int f()

...{

return 0;

}

test1:

>gcc -c a.c -o a.o

>nm -s a.o

00000030 T f
00000000 t gcc2_compiled.
00000000 T main
U printf

>gcc -c b.c -o b.o

>nm -s b.o

00000000 T f
00000000 t gcc2_compiled.

>gcc b.o a.o -o main //link failed because of the same global symbols "f"

a.o: In function `f':
a.o(.text+0x30): multiple definition of `f'
b.o(.text+0x0): first defined here
collect2: ld returned 1 exit status

test2:

>ar cr b.a b.o

f in b.o

b.o:
00000000 T f
00000000 t gcc2_compiled.

>ar cr a.a a.o

Archive index:
main in a.o
f in a.o

a.o:
00000030 T f
00000000 t gcc2_compiled.
00000000 T main
U printf

>gcc b.a a.a -o main //link is sucessful!!! what's the difference comparing with test1???
> main

result is 1 //why is 1???

--------------------------------------------------------------------------------------------------------------------------------------------

a1.c:

extern int f();

int main()

...{

printf("result is %d ",f());

return f();

}

a2.c:

int f()

...{

return 1;

}

b.c:

int f()

...{

return 0;

}

> gcc -c a1.c -o a1.o

U f
00000000 t gcc2_compiled.
00000000 T main
U printf

> gcc -c a2.c -o a2.o

00000000 T f
00000000 t gcc2_compiled.

> gcc -c b.c -o b.o

00000000 T f
00000000 t gcc2_compiled.
> ar cr a2.a a2.o

> nm -s a2.a

Archive index:
f in a2.o

a2.o:
00000000 T f
00000000 t gcc2_compiled.
> ar cr b.a b.o

> nm -s b.a

Archive index:
f in b.o

b.o:
00000000 T f
00000000 t gcc2_compiled.

> ar cr a1.a a1.o

> nm -s a1.a

Archive index:
main in a1.o

a1.o:
U f
00000000 t gcc2_compiled.
00000000 T main
U printf

> ar cr a.a a2.o a1.o

> nm -s a.a

Archive index:
f in a2.o
main in a1.o

a2.o:
00000000 T f
00000000 t gcc2_compiled.

a1.o:
U f
00000000 t gcc2_compiled.
00000000 T main
U printf

test1:
> gcc b.a a.a -o main

> main

result is 1

test2:

> gcc a1.a b.a a2.a -o main0

> gcc a2.a a1.a b.a -o main0

> main0

result is 0 //the result depends on linking sequence!!!

> gcc a1.a a2.a b.a -o main1

> gcc b.a a1.a a2.a -o main1

> main1

result is 1 //the result depends on linking sequence!!!

test3:

> gcc b.a a2.a a1.a -main //why failed???

a1.a(a1.o): In function `main':
a1.o(.text+0xd): undefined reference to `f'
a1.o(.text+0x25): undefined reference to `f'
collect2: ld returned 1 exit status