C++ 共享库初始化全局变量

Question

shared.h

    #pragma once
    
    extern int uninitialized_variable;
    extern int initialized_variable;
    
    void print();

shared.cpp

    // shared.cpp
    #include "shared.h"
    #include <stdio.h>
    
    int uninitialized_variable;
    int initialized_variable = 8;
    
    void print() {
        printf("%d\n", uninitialized_variable);
        printf("%d\n", initialized_variable);
    }

main.cpp

    #include "shared.h"
    #include <stdio.h>
    
    
    int main() {
        printf("%d\n", uninitialized_variable);
        printf("%d\n", initialized_variable);
        print();
    }

g++ -O0 -pedantic -std=c++11 -fPIC shared.cpp -c -o shared.o
g++ -shared -o libshared.so shared.o

g++ -O0 -pedantic -std=c++11 main.cpp -L. -lshared

When I run./a.out, the result is 0 8 0 8. But I can't find the reason why initialized_variable is 8.

Since I used gbd the check where exactly is initialized_variable ，and found that it is in the.bss in a.out。But in the.init section of the libshared.so, i can't find any code to do the initialization of the initialized_variable in a.out's.bss.

Who and when the initialized_variable is initalized?

Answer 1

It's provided as raw data in the .data segment:

Disassembly of section .data:

....

0000000000004028 <initialized_variable>:
    4028:   08 00                   or     %al,(%rax)
    402a:   00 00                   add    %al,(%rax)

Ignore the assembly as this is not executable data. Note that the offset where initialized_variable lives contains the four bytes (in hex here): 08 00 00 00. Interpreted as a 4-byte signed little-endian integer, this is the value 8.

Quite simply, the value is 8 as soon as the library is mapped into memory. No additional instructions are required to initialize it.