3

Hi i want to make a function which calculate the size of structure without padding bytes.

Example :

struct test{
    int   x;
    char  y;
    int   z;
};

If i calculate the size of above structure i must get 9 byte (i.e. without padding bytes)

And consider the case where i might not know the variable present in struct.

Example :

struct test{
    int   x;
    ....
    ....
    ....
    int   z;
};

so if i calculate the size it should give correct size.

Is it possible do write such function?I tried reading on structure's but i dint find any solution.I saw there is some compiler option are present from that i can get but i dont want any in build compiler option.

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4
  • You can always add your own size function or use sizeof with some special compiler specific magic. Commented Feb 4, 2013 at 16:48
  • 1
    #pragma pack(1) to force it down to unpadded size? Commented Feb 4, 2013 at 16:48
  • @Dukeling i know through standard options i will get as i mentioned in question, but i dont need such options. Commented Feb 4, 2013 at 16:50
  • 1
    Why? There's nothing useful that you can do with such information. Commented Feb 4, 2013 at 17:53

2 Answers 2

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4

No, C doesn't have enough introspection to make this possible. At run-time, there is no information left that the program can use to "know" which fields are in a structure, or what their types are.

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4 Comments

ok that valid for case 2.But suppose if we know the the varibles of struct?
Pass each variable to sizeof() and sum the sizes
@Badshah: it's not possible either. What you would need is to generically iterate over all the member variables of a class and apply sizeof to each, then do the sum. But there's no way to perform such an iteration
@AndyProwl yes i know even if i will do also somehow it will be a bulky process.
3

The sizeof compiler construct should give you the size of the object "as the compiler sees it". Most compilers will align structures on 4 byte boundaries for efficient memory access at the expense of space. Try

#include <stdio.h>
int main() {
  struct test{
    int x;
    char a;
    int y;
  };
  struct test myStruct;
  printf("size of myStruct is %ld\n", sizeof(myStruct));
}

When I run it, I get

size of myStruct is 12

Which shows that my compiler, in this situation, is allocating four bytes to 'a'. I think that's your answer... unless I really misunderstood your question, and in particular your "without padding bytes" comment.

Rewriting the definition of the struct as:

  struct test{
    int x;
    char a;
    int y;
  }__attribute__((packed));

The output becomes

size of myStruct is 9

This only works at compile time - so it is not really a "function". And it doesn't apply if you don't know your struct at compile time. As such, I think my answer and the one given by @unwind above are not inconsistent.

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4 Comments

above will give me size of overall struct i.e. 12 bytes not 9 bytes.
size of operator will give the overall structure size i.e.(int 4 , char 1 (3 bytes paaded ) 1 + 3 = 4, int 4, so total - 4 + 4 + 4 = 12.But i need actuall i.e. 9 bytes ...
yes now it will give 9 because we are using pragma "__attribute__((packed));"
Actually, in most compilers, the alignment of a struct will vary according to its contents. Something like struct X { char a; char b; char c; } will probably have an alignment of 1, and a size of 3; if the struct contains a double, it will usually have an alignment of 8.

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