问题描述
我试图学习C程序的可执行文件的结构。我的环境是GCC和64位英特尔处理器。考虑以下C代码 a.cc 。
#include< cstdlib>
#include< cstdio>
int x;
int main（）{
printf（％d \ n，sizeof（x））;
返回10;
大小-oa 显示
文本数据bss十六进制文件名
1134 552 8 1694 69e a
在我添加了另一个初始化的全局变量y之后。
int y = 10;
size a 显示（其中 a 是来自a.cc的可执行文件的名称）
文本数据bss dec hex文件名
1134 556 12 1702 6a6 a
我们知道， BSS 节存储未初始化的全局变量的大小， DATA 存储已初始化的全局变量。
- 为什么 int 在BSS中占用8个字节？我的代码中的 sizeof（x）显示 int 实际占用了4个字节。
- int y = 10 将4个字节添加到DATA，这是有意义的，因为 int 应占用4个字节。但是，为什么它会向BSS添加4个字节？
- Why int takes up 8 bytes in BSS? The sizeof(x) in my code shows that the int actually takes up 4 bytes.
- The int y=10 added 4 bytes to DATA which makes sense since int should take 4 bytes. But, why does it adds 4 bytes to BSS?
两个大小命令在删除两行 #include ... 后保持不变。
更新：
我认为我对BSS的理解是错误的。它可能不会存储未初始化的全局变量。正如维基百科所说：BSS在运行时需要的大小被记录在目标文件中，但是BSS（与数据段不同）不会占用目标文件中的任何实际空间。例如，即使是一行C代码 int main（）{} 也有 bss 8 。
BSS的8个或16个来自对齐？
它不会占用4个字节，无论它在哪个分段。您可以使用 nm 工具（来自GNU binutils包）与 -S 参数进行比较，以获取目标文件中所有符号的名称和大小。您可能会看到编译器的次要影响，包括或不包括某些其他符号的原因。
$ cat a1.c
int x;
$ cat a2.c
int x = 1;
$ gcc -c a1.c a2.c
$ nm -S a1.o a2.o
a1.o：
0000000000000004 0000000000000004 C x
a2.o：
0000000000000000 0000000000000004 D x
一个目标文件在未初始化的数据段（ C ）中有一个名为 x 的4字节对象，而另一个对象文件具有在初始化数据段（ D ）中一个名为 x 的4字节对象。
I am trying to learn the structure of executable files of C program. My environment is GCC and 64bit Intel processor.
Consider the following C code a.cc.
#include <cstdlib>
#include <cstdio>
int x;
int main(){
printf("%d\n", sizeof(x));
return 10;
}
The size -o a shows
text data bss dec hex filename 1134 552 8 1694 69e a
After I added another initialized global variable y.
int y=10;
The size a shows (where a is the name of the executable file from a.cc)
text data bss dec hex filename 1134 556 12 1702 6a6 a
As we know, the BSS section stores the size of uninitialized global variables and DATA stores initialized ones.
The difference between two size commands stays the same after deleting the two lines #include ....
Update:I think my understanding of BSS is wrong. It may not store the uninitialized global variables. As the Wikipedia says "The size that BSS will require at runtime is recorded in the object file, but BSS (unlike the data segment) doesn't take up any actual space in the object file." For example, even the one line C code int main(){} has bss 8.
Does the 8 or 16 of BSS comes from alignment?
It doesn't, it takes up 4 bytes regardless of which segment it's in. You can use the nm tool (from the GNU binutils package) with the -S argument to get the names and sizes of all of the symbols in the object file. You're likely seeing secondary affects of the compiler including or not including certain other symbols for whatever reasons.
For example:
$ cat a1.c int x; $ cat a2.c int x = 1; $ gcc -c a1.c a2.c $ nm -S a1.o a2.o a1.o: 0000000000000004 0000000000000004 C x a2.o: 0000000000000000 0000000000000004 D x
One object file has a 4-byte object named x in the uninitialized data segment (C), while the other object file has a 4-byte object named x in the initialized data segment (D).
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