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Home/ Questions/Q 8795527
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T23:24:23+00:00 2026-06-13T23:24:23+00:00

I’m reading in the first Bytes of an File with fread: fread(&example_struct, sizeof(example_struct), 1,

  • 0

I’m reading in the first Bytes of an File with fread:

fread(&example_struct, sizeof(example_struct), 1, fp_input);

Which ends up with different results under linux and solaris? Whereby the example_struct (Elf32_Ehdr) is part of Standart GNU C Liborary defined in elf.h? I would be happy to know why this happens?

General the struct looks the following:

typedef struct
{
  unsigned char e_ident[LENGTH];    
  TYPE_Half e_type;         
} example_struct;

The Debugcode:

for(i=0;paul<sizeof(example_struct);i++){
    printf("example_struct->e_ident[%i]:(%x) \n",i,example_struct.e_ident[i]);
}
printf("example_struct->e_type: (%x) \n",example_struct.e_type);
printf("example_struct->e_machine: (%x) \n",example_struct.e_machine);

Solaris output:

Elf32_Ehead->e_ident[0]: (7f) 
Elf32_Ehead->e_ident[1]: (45) 
...
Elf32_Ehead->e_ident[16]: (2) 
Elf32_Ehead->e_ident[17]: (0)
...
Elf32_Ehead->e_type: (200) 
Elf32_Ehead->e_machine: (6900) 

Linux output:

Elf32_Ehead->e_ident[0]: (7f) 
Elf32_Ehead->e_ident[1]: (45) 
...
Elf32_Ehead->e_ident[16]: (2) 
Elf32_Ehead->e_ident[17]: (0)
...
Elf32_Ehead->e_type: (2) 
Elf32_Ehead->e_machine: (69)

Maybe similar to: http://forums.devarticles.com/c-c-help-52/file-io-linux-and-solaris-108308.html

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T23:24:25+00:00Added an answer on June 13, 2026 at 11:24 pm

    You don’t mention what CPU you have in the machines, maybe Sparc64 in the Solaris machine and x86_64 in the Linux box, but I would guess that you’re having an endianness issue. Intel, ARM and most other common architectures today are what is known as little-endian, the Sparc architecture is big-endian.

    Let’s assume we have the value 0x1234 in a CPU register and we want to store it in memory (or on hard drive, it doesn’t matter where). Let N be the memory address we want to write to. We will need to store this 16 bit integer as two bytes in memory, here comes the confusing part:

    Using a big-endian machine will store 0x12 at address N and 0x34 at address N+1.
    A little-endian machine will store 0x34 at address N and 0x12 at address N+1.
    If we store a value using a little endian machine and read it back using a big endian machine we will have swapped the two bytes around and you’ll get the issue that you are seeing.

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