I am facing issue with c library function strtoull which is returning me wrong output.
int main(int argc, char *argv[])
{
unsigned long long int intValue;
if(atoi(argv[2]) == 1)
{
intValue = strtoull((const char *)argv[1], 0, 10 );
}
else
{
// ...
}
printf("intValue of %s is %llu \n", argv[1], intValue);
return 0;
}
I built them and generated 32 and 64 bit executables as str32_new and str64_new.
But the output received from 32 bit exe is errorneous as wrong number is returned:
strtoull should had returned me number 5368709120 for the passed string "5368709120" but it returned me 1073741824.
# ./str32_new "5368709120" 1
intValue of 5368709120 is 1073741824
I note that when I decrease one character from string then it shows proper output.
# ./str32_new "536870912" 1
intValue of 536870912 is 536870912
glibc attached to 32 bit exe is
# readelf -Wa /home/str32_new | grep strt
[39] .shstrtab STRTAB 00000000 002545 000190 00 0 0 1
[41] .strtab STRTAB 00000000 0032f8 0002a4 00 0 0 1
0804a014 00000607 R_386_JUMP_SLOT 00000000 strtoull
6: 00000000 0 FUNC GLOBAL DEFAULT UND strtoull#GLIBC_2.0 (2)
55: 00000000 0 FILE LOCAL DEFAULT ABS strtoull.c
75: 00000000 0 FUNC GLOBAL DEFAULT UND strtoull##GLIBC_2.0
77: 08048534 915 FUNC GLOBAL DEFAULT 15 my_strtoull
glibc attached to 64 bit exe is
# readelf -Wa /home/str64_new | grep strt
[39] .shstrtab STRTAB 0000000000000000 001893 000192 00 0 0 1
[41] .strtab STRTAB 0000000000000000 002cd0 00029b 00 0 0 1
0000000000601028 0000000700000007 R_X86_64_JUMP_SLOT 0000000000000000 strtoull + 0
7: 0000000000000000 0 FUNC GLOBAL DEFAULT UND strtoull#GLIBC_2.2.5 (2)
57: 0000000000000000 0 FILE LOCAL DEFAULT ABS strtoull.c
73: 00000000004006cc 804 FUNC GLOBAL DEFAULT 15 my_strtoull
82: 0000000000000000 0 FUNC GLOBAL DEFAULT UND strtoull##GLIBC_2.2.5
64 bit exe shows proper output but on some system it too behaves abnormally.
Why is the strtoull in 32 bit exe behaving so and how to resolve this issue?
Ok, so we've established that this is quite obviously happening due to an overflow, as the value matches what would happen if casted into 32bit int.
This however does not explain everything - you did use strtoull, not the shorter strtoul, and it indeed works on 64bit binary. If anything, I was surprised to see you were even able to call the longer version in your 32bit build (how did you build it by the way, with -m32? or on a special machine?)
This link, raises the possibility that there's some linkage phenomenon that makes strtoull get declared as int strtoll() (presumably the system can't support the original lib version), and so we get the value implicitly casted through int, before copied back to your unsigned long long.
Either way - this should have been warned against by the compiler, try setting it to c99 and raise the warning levels, maybe that would make it shout
I think that is due to overflow. int in 32bit can not hold a number that large (max is 4294967296). As Leeor said, 5368709120 & (0xffffffff) = 1073741824.
The type int is minimally 32 bits wide, and is only 32 bits wide on most (if not all) systems.
You most likely forgot to #include <stdlib.h> and you probably did not enable any compiler warnings (like for using undeclared functions).
When a C compiler sees a function call to an undeclared function it blindly assumes int f(int) as prototype. In your case the return value of strtoull() will be int and so the value will be truncated to 32-bit.
(It is indeed quite strange that you get the correct result on a 64-bit system, where int is usually also just 32-bit.)
Related
I'm on an Ubuntu 18.04 laptop coding C with VSCode and compiling it with GNU's gcc.
I'm doing some basic engineering on my own C code and I noticed a few interesting details, on of which is the pair []A\A]A^A_ and ;*3$" that seems to appear in every one of my compiled C binaries. Between them is usually (or always) strings that I hard code in for printf() functions.
An example is this short piece of code here:
#include <stdio.h>
#include <stdbool.h>
int f(int i);
int main()
{
int x = 5;
int o = f(x);
printf("The factorial of %d is: %d\n", x, o);
return 0;
}
int f(int i)
{
if(i == 0)
{
return i;
}
else
{
return i*f(i-1);
}
}
... is then compiled using gcc test.c -o test.
When I run strings test, the following is outputted:
/lib64/ld-linux-x86-64.so.2
0HSn(
libc.so.6
printf
__cxa_finalize
__libc_start_main
GLIBC_2.2.5
_ITM_deregisterTMCloneTable
__gmon_start__
_ITM_registerTMCloneTable
AWAVI
AUATL
[]A\A]A^A_
The factorial of %d is: %d
;*3$"
GCC: (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0
crtstuff.c
deregister_tm_clones
__do_global_dtors_aux
completed.7697
__do_global_dtors_aux_fini_array_entry
frame_dummy
__frame_dummy_init_array_entry
test.c
__FRAME_END__
__init_array_end
_DYNAMIC
__init_array_start
__GNU_EH_FRAME_HDR
_GLOBAL_OFFSET_TABLE_
__libc_csu_fini
_ITM_deregisterTMCloneTable
_edata
printf##GLIBC_2.2.5
__libc_start_main##GLIBC_2.2.5
__data_start
__gmon_start__
__dso_handle
_IO_stdin_used
__libc_csu_init
__bss_start
main
__TMC_END__
_ITM_registerTMCloneTable
__cxa_finalize##GLIBC_2.2.5
.symtab
.strtab
.shstrtab
.interp
.note.ABI-tag
.note.gnu.build-id
.gnu.hash
.dynsym
.dynstr
.gnu.version
.gnu.version_r
.rela.dyn
.rela.plt
.init
.plt.got
.text
.fini
.rodata
.eh_frame_hdr
.eh_frame
.init_array
.fini_array
.dynamic
.data
.bss
.comment
Same as other scripts I've written, the 2 pieces []A\A]A^A_ and ;*3$" always pop up, 1 before the strings used with printf and one right after.
I'm curious: What exactly do those strings mean? I'm guessing they mainly mark the begining and endding of the use of hard-coded output strings.
Our digital computers work on bits, most commonly clustered in bytes containing 8 bits each. The meaning of such a combination depends on the context and the interpretation.
A non-exhausting list of possible interpretation is:
ASCII characters with the eighth bit ignored or accepted only if 0;
signed or unsigned 8-bit integer;
operation code (or part of it) of one specific machine language, each processor (family) has its own different set.
For example, the hex value 0x43 can be seen as:
ASCII character 'C';
Unsigned 8-bit integer 67 (signed is the same if 2's complement is used);
Operation code "LD B,E" for a Z80 CPU (see, I'm really old and learned that processor in depth);
Operation code "EORS ari" for an ARM CPU.
Now strings simply (not to say "primitively") scans through the given file and tries so interpret the bytes as sequences of printable ASCII characters. By default a sequence has to have at least 4 characters and the bytes are interpreted as 7-bit ASCII. BTW, the file does not have to be an executable. You can scan any file but if you give it an object file by default it scans only sections that are loaded in memory.
So what you see are sequences of bytes which by chance are at least 4 printable characters in a row. And because some patterns are always in an executable it just looks as if they have a special meaning. Actually they have but they don't have to relate to your program's strings.
You can use strings to quickly peek into a file to find, well, strings which might help you with whatever you're trying to accomplish.
What you're seeing is an ASCII representation of a particular bit pattern that happens to be common in executable programs generated by that particular compiler. The pattern might correspond to a particular sequence of machine language instructions which the compiler is fond of emitting. Or it might correspond to a particular data structure which the compiler or linker uses to mark the various other pieces of data stored in the executable.
Given enough work, it would probably be possible to work out the actual details, for your C code and your particular version of your particular compiler, precisely what the bit patterns behind []A\A]A^A_ and ;*3$" correspond to. But I don't do much machine-language programming any more, so I'm not going to try, and the answers probably wouldn't be too interesting in the end, anyway.
But it reminds me of little quirk which I have noticed and can explain. Suppose you wrote the very simple program
int i = 12345;
If you compiled that program and ran strings on it, and if you told it to look for strings as short as two characters, you'd probably see (among lots of other short, meaningless strings), the string
90
and that bit pattern would, in fact, correspond to your variable! What's up with that?
Well, 12345 in hexadecimal is 0x3039, and most machines these days are little-endian, so those two bytes in memory are stored in the other order as
39 30
and in ASCII, 0x39 is '9', while 0x30 is '0'.
And if this is interesting to you, you can try compiling the program fragment
int i = 12345;
long int a = 1936287860;
long int b = 1629516649;
long int c = 1953719668;
long long int x = 48857072035144;
long long int y = 36715199885175;
and running strings -2 on it, and see what else you get.
When checking the disassembly of the object file through the readelf, I see the data and the bss segments contain the same offset address.
The data section will contain the initialized global and static variables. BSS will contain un-initialized global and static variables.
1 #include<stdio.h>
2
3 static void display(int i, int* ptr);
4
5 int main(){
6 int x = 5;
7 int* xptr = &x;
8 printf("\n In main() program! \n");
9 printf("\n x address : 0x%x x value : %d \n",(unsigned int)&x,x);
10 printf("\n xptr points to : 0x%x xptr value : %d \n",(unsigned int)xptr,*xptr);
11 display(x,xptr);
12 return 0;
13 }
14
15 void display(int y,int* yptr){
16 char var[7] = "ABCDEF";
17 printf("\n In display() function \n");
18 printf("\n y value : %d y address : 0x%x \n",y,(unsigned int)&y);
19 printf("\n yptr points to : 0x%x yptr value : %d \n",(unsigned int)yptr,*yptr);
20 }
output:
SSS:~$ size a.out
text data bss dec hex filename
1311 260 8 1579 62b a.out
Here in the above program I don't have any un-initialized data but the BSS has occupied 8 bytes. Why does it occupy 8 bytes?
Also when I disassemble the object file,
EDITED :
[ 3] .data PROGBITS 00000000 000110 000000 00 WA 0 0 4
[ 4] .bss NOBITS 00000000 000110 000000 00 WA 0 0 4
[ 5] .rodata PROGBITS 00000000 000110 0000cf 00 A 0 0 4
data, rodata and bss has the same offset address. Does it mean the rodata, data and bss refer to the same address?
Do Data section, rodata section and the bss section contain the data values in the same address, if so how to distinguish the data section, bss section and rodata section?
The .bss section is guaranteed to be all zeros when the program is loaded into memory. So any global data that is uninitialized, or initialized to zero is placed in the .bss section. For example:
static int g_myGlobal = 0; // <--- in .bss section
The nice part about this is, the .bss section data doesn't have to be included in the ELF file on disk (ie. there isn't a whole region of zeros in the file for the .bss section). Instead, the loader knows from the section headers how much to allocate for the .bss section, and simply zero it out before handing control over to your program.
Notice the readelf output:
[ 3] .data PROGBITS 00000000 000110 000000 00 WA 0 0 4
[ 4] .bss NOBITS 00000000 000110 000000 00 WA 0 0 4
.data is marked as PROGBITS. That means there are "bits" of program data in the ELF file that the loader needs to read out into memory for you. .bss on the other hand is marked NOBITS, meaning there's nothing in the file that needs to be read into memory as part of the load.
Example:
// bss.c
static int g_myGlobal = 0;
int main(int argc, char** argv)
{
return 0;
}
Compile it with $ gcc -m32 -Xlinker -Map=bss.map -o bss bss.c
Look at the section headers with $ readelf -S bss
Section Headers:
[Nr] Name Type Addr Off Size ES Flg Lk Inf Al
[ 0] NULL 00000000 000000 000000 00 0 0 0
:
[13] .text PROGBITS 080482d0 0002d0 000174 00 AX 0 0 16
:
[24] .data PROGBITS 0804964c 00064c 000004 00 WA 0 0 4
[25] .bss NOBITS 08049650 000650 000008 00 WA 0 0 4
:
Now we look for our variable in the symbol table: $ readelf -s bss | grep g_myGlobal
37: 08049654 4 OBJECT LOCAL DEFAULT 25 g_myGlobal
Note that g_myGlobal is shown to be a part of section 25. If we look back in the section headers, we see that 25 is .bss.
To answer your real question:
Here in the above program I dont have any un-intialised data but the BSS has occupied 8 bytes. Why does it occupy 8 bytes ?
Continuing with my example, we look for any symbol in section 25:
$ readelf -s bss | grep 25
9: 0804825c 0 SECTION LOCAL DEFAULT 9
25: 08049650 0 SECTION LOCAL DEFAULT 25
32: 08049650 1 OBJECT LOCAL DEFAULT 25 completed.5745
37: 08049654 4 OBJECT LOCAL DEFAULT 25 g_myGlobal
The third column is the size. We see our expected 4-byte g_myGlobal, and this 1-byte completed.5745. This is probably a function-static variable from somewhere in the C runtime initialization - remember, a lot of "stuff" happens before main() is ever called.
4+1=5 bytes. However, if we look back at the .bss section header, we see the last column Al is 4. That is the section alignment, meaning this section, when loaded, will always be a multiple of 4 bytes. The next multiple up from 5 is 8, and that's why the .bss section is 8 bytes.
Additionally We can look at the map file generated by the linker to see what object files got placed where in the final output.
.bss 0x0000000008049650 0x8
*(.dynbss)
.dynbss 0x0000000000000000 0x0 /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../lib/crt1.o
*(.bss .bss.* .gnu.linkonce.b.*)
.bss 0x0000000008049650 0x0 /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../lib/crt1.o
.bss 0x0000000008049650 0x0 /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../lib/crti.o
.bss 0x0000000008049650 0x1 /usr/lib/gcc/x86_64-redhat-linux/4.7.2/32/crtbegin.o
.bss 0x0000000008049654 0x4 /tmp/ccKF6q1g.o
.bss 0x0000000008049658 0x0 /usr/lib/libc_nonshared.a(elf-init.oS)
.bss 0x0000000008049658 0x0 /usr/lib/gcc/x86_64-redhat-linux/4.7.2/32/crtend.o
.bss 0x0000000008049658 0x0 /usr/lib/gcc/x86_64-redhat-linux/4.7.2/../../../../lib/crtn.o
Again, the third column is the size.
We see 4 bytes of .bss came from /tmp/ccKF6q1g.o. In this trivial example, we know that is the temporary object file from the compilation of our bss.c file. The other 1 byte came from crtbegin.o, which is part of the C runtime.
Finally, since we know that this 1 byte mystery bss variable is from crtbegin.o, and it's named completed.xxxx, it's real name is completed and it's probably a static inside some function. Looking at crtstuff.c we find the culprit: a static _Bool completed inside of __do_global_dtors_aux().
By definition, the bss segment takes some place in memory (when the program starts) but don't need any disk space. You need to define some variable to get it filled, so try
int bigvar_in_bss[16300];
int var_in_data[5] = {1,2,3,4,5};
Your simple program might not have any data in .bss, and shared libraries (like libc.so) may have "their own .bss"
File offsets and memory addresses are not easily related.
Read more about the ELF specification, also use /proc/ (eg cat /proc/self/maps would display the address space of the cat process running that command).
Read also proc(5)
Is it mandatory to have unique names for every program section in bpf program? For instance, this program compiles fine with llvm-5.0 :
...
SEC("sockops")
int bpf1(struct bpf_sock_ops *sk_ops)
{
return 1;
}
SEC("sockops")
int bpf2(struct bpf_sock_ops *sk_ops)
{
return 1;
}
SEC("sockops")
int bpf3(struct bpf_sock_ops *sk_ops)
{
return 1;
}
SEC("sockops")
int bpf_main(struct bpf_sock_ops *sk_ops)
{
__u32 port = bpf_ntohl(sk_ops->remote_port);
switch (port) {
case 5000:
bpf_tail_call(sk_ops, &jmp_table, 1);
break;
case 6000:
bpf_tail_call(sk_ops, &jmp_table, 2);
break;
case 7000:
bpf_tail_call(sk_ops, &jmp_table, 3);
break;
}
sk_ops->reply = 0;
return 1;
}
char _license[] SEC("license") = "GPL";
u32 _version SEC("version") = LINUX_VERSION_CODE;
However, llvm-objdump reports only one program section:
$ llvm-objdump-5.0 -section-headers bpf_main.o
bpf_main.o: file format ELF64-BPF
Sections:
Idx Name Size Address Type
0 00000000 0000000000000000
1 .strtab 000000a5 0000000000000000
2 .text 00000000 0000000000000000 TEXT DATA
3 sockops 000001f8 0000000000000000 TEXT DATA
4 .relsockops 00000030 0000000000000000
5 maps 0000001c 0000000000000000 DATA
6 .rodata.str1.16 00000021 0000000000000000 DATA
7 .rodata.str1.1 0000000e 0000000000000000 DATA
8 license 00000004 0000000000000000 DATA
9 version 00000004 0000000000000000 DATA
10 .eh_frame 00000090 0000000000000000 DATA
11 .rel.eh_frame 00000040 0000000000000000
12 .symtab 00000138 0000000000000000
Is there a compiler option to give at least a warning?
UPDATE - so I understand that sections provided in the same bpf file must have unique names. Is it true in cases when bpf programs reside in different files and compiled independently, so that they could be loaded and call one another, e.g. tail calls?
Apparently the compiler concatenates everything in the same section (I simplified bpf_main to have it compile and to test).
$ readelf -x sockops bpf_prog.o
Hex dump of section 'sockops':
0x00000000 b7000000 01000000 95000000 00000000 ................
0x00000010 b7000000 01000000 95000000 00000000 ................
0x00000020 b7000000 01000000 95000000 00000000 ................
0x00000030 b7000000 00000000 95000000 00000000 ...............
From here, I cannot see how you intend to later retrieve the individual programs to load them properly. Usually, user space tools get one program from a section, and attempt to load them by passing the instructions to the kernel through the bpf() syscall; here, whatever program you use will probably try to load the concatenated four programs at once.
Is there a particular reason why you would like to have all programs in sections with the same name?
I have no idea for the warning in LLVM on such a thing. I guess not: people may have reason to put different stuff in a section. The case here is specific to BPF, and I do not think people usually try to put several programs in a same section. But this is just a guess, I may be wrong.
Regarding your update:
I do not believe using similar sections names in different object files is an issue. As long as your user space tool is able to retrieve the programs from object files and to perform relocation, it should work fine. The kernel does not see any section name anyway. What it takes from the attr argument is an array of instructions:
struct { /* anonymous struct used by BPF_PROG_LOAD command */
__u32 prog_type; /* one of enum bpf_prog_type */
__u32 insn_cnt;
__aligned_u64 insns;
__aligned_u64 license;
__u32 log_level; /* verbosity level of verifier */
__u32 log_size; /* size of user buffer */
__aligned_u64 log_buf; /* user supplied buffer */
__u32 kern_version; /* checked when prog_type=kprobe */
__u32 prog_flags;
char prog_name[BPF_OBJ_NAME_LEN];
__u32 prog_ifindex; /* ifindex of netdev to prep for */
};
(from include/uapi/linux/bpf.h, note the insns attribute). For tail calls your user space code also needs to create the specific maps holding referenced to the programs you want to jump to (more bpf() calls). But there again, it's all the responsibility of user space to extract the information from the ELF files. Libbpf should be able to handle it correctly I think, but I have not tried it.
A side note to finish: I don't know what your use case is exactly, but since you are trying to compile your code in different object files and to use calls, you might be interested to learn that eBPF now supports “function calls”, i.e. you can define several (non-inline) functions, possibly in several ELF files, and call them in your program. More information in the cover letter. Again, I didn't have the time to experiment with it so far.
is vdso supported for a 32 bit application which is running on a 64 bit kernel with glibc version 2.15.? If yes, how do I make it work for 32 bit application running on 64 bit kernel.? Cause even though dlopen on "linux-vdso.so.1" is success, dlsym on "__vdso_gettimeofday" fails.
On the same system I able to do a dlopen on "linux-vdso.so.1" & dlsym on "__vdso_gettimeofday" from a application compiled for 64 bit.
On my 64-bit Linux 4.4.15, the 32-bit vdso has these symbols:
readelf -Ws vdso32
Symbol table '.dynsym' contains 9 entries:
Num: Value Size Type Bind Vis Ndx Name
0: 00000000 0 NOTYPE LOCAL DEFAULT UND
1: 00000ce0 9 FUNC GLOBAL DEFAULT 12 __kernel_sigreturn##LINUX_2.5
2: 00000d00 13 FUNC GLOBAL DEFAULT 12 __kernel_vsyscall##LINUX_2.5
3: 00000ad0 438 FUNC GLOBAL DEFAULT 12 __vdso_gettimeofday##LINUX_2.6
4: 00000c90 42 FUNC GLOBAL DEFAULT 12 __vdso_time##LINUX_2.6
5: 00000770 853 FUNC GLOBAL DEFAULT 12 __vdso_clock_gettime##LINUX_2.6
6: 00000cf0 8 FUNC GLOBAL DEFAULT 12 __kernel_rt_sigreturn##LINUX_2.5
7: 00000000 0 OBJECT GLOBAL DEFAULT ABS LINUX_2.5
8: 00000000 0 OBJECT GLOBAL DEFAULT ABS LINUX_2.6
This suggests that the __vdso_gettimeofday you are looking for has been added in kernel 2.6, and that your kernel version is older.
For some reason I made simple program in C to output binary representation of given input:
int main()
{
char c;
while(read(0,&c,1) > 0)
{
unsigned char cmp = 128;
while(cmp)
{
if(c & cmp)
write(1,"1",1);
else
write(1,"0",1);
cmp >>= 1;
}
}
return 0;
}
After compilation:
$ gcc bindump.c -o bindump
I made simple test to check if program is able to print binary:
$ cat bindump | ./bindump | fold -b100 | nl
Output is following: http://pastebin.com/u7SasKDJ
I suspected the output to look like random series of ones and zeroes. However, output partially seems to be quite more interesting. For example take a look at the output between line 171 and 357. I wonder why there are lots of zeros in compare to other sections of executable ?
My architecture is:
$ lscpu
Architecture: i686
CPU op-mode(s): 32-bit, 64-bit
Byte Order: Little Endian
CPU(s): 4
On-line CPU(s) list: 0-3
Thread(s) per core: 2
Core(s) per socket: 2
Socket(s): 1
Vendor ID: GenuineIntel
CPU family: 6
Model: 28
Stepping: 10
CPU MHz: 1000.000
BogoMIPS: 3325.21
Virtualization: VT-x
L1d cache: 24K
L1i cache: 32K
L2 cache: 512K
When you compile a program into an executable on Linux (and a number of other unix systems), it is written in the ELF format. The ELF format has a number of sections, which you can examine with readelf or objdump:
readelf -a bindump | less
For example, section .text contains CPU instructions, .data global variables, .bss uninitialized global variables (it is actually empty in the ELF file itself, but is created in the main memory when the program is executed), .plt and .got which are jump tables, debugging information, etc.
Btw. it is much more convenient to examine the binary content of files with hexdump:
hexdump -C bindata | less
There you can see that starting with offset 0x850 (approx. line 171 in your dump) there is a lot of zeros, and you can also see the ASCII representation on the right.
Let us look at which sections correspond to the block of your interest between 0x850 and 0x1160 (the field Off – offset in the file is important here):
> readelf -a bindata
...
Section Headers:
[Nr] Name Type Addr Off Size ES Flg Lk Inf Al
...
[28] .shstrtab STRTAB 00000000 00074c 000106 00 0 0 1
[29] .symtab SYMTAB 00000000 000d2c 000440 10 30 45 4
...
You can examine the content of an individual section with -x:
> readelf -x .symtab bindump | less
0x00000000 00000000 00000000 00000000 00000000 ................
0x00000010 00000000 34810408 00000000 03000100 ....4...........
0x00000020 00000000 48810408 00000000 03000200 ....H...........
0x00000030 00000000 68810408 00000000 03000300 ....h...........
0x00000040 00000000 8c810408 00000000 03000400 ................
0x00000050 00000000 b8810408 00000000 03000500 ................
0x00000060 00000000 d8810408 00000000 03000600 ................
You would see that there are many zeros. The section is composed of 18-byte values (= one line in the -x output) defining symbols. From readelf -a you can see that it has 68 entries, and first 27 of them (excl. the very first one) are of type SECTION:
Symbol table '.symtab' contains 68 entries:
Num: Value Size Type Bind Vis Ndx Name
0: 00000000 0 NOTYPE LOCAL DEFAULT UND
1: 08048134 0 SECTION LOCAL DEFAULT 1
2: 08048148 0 SECTION LOCAL DEFAULT 2
3: 08048168 0 SECTION LOCAL DEFAULT 3
4: 0804818c 0 SECTION LOCAL DEFAULT 4
...
According to the specification (page 1-18), each entry has the following format:
typedef struct {
Elf32_Word st_name;
Elf32_Addr st_value;
Elf32_Word st_size;
unsigned char st_info;
unsigned char st_other;
Elf32_Half st_shndx;
} Elf32_Sym;
Without going into too much detail here, I think what matters here is that st_name and st_size are both zeros for these SECTION entries. Both are 32-bit numbers, which means lots of zeros in this particular section.
This is not really a programming question, but however...
A binary normally consists of different sections: code, data, debugging info, etc. Since these sections contents differ by type, I would pretty much expect them to look different.
I.e. the symbol table consists of address offsets in your binary. If I read your lspci correctly, you are on a 32-bit system. That means Each offset has four bytes, and given the size of your program, in most cases two of those bytes will be zero. And there are more effects like this.
You didn't strip your program, that means there's still lots of information (symbol table etc.) present in the binary. Try stripping the binary and have a look at it again.