C Struct Packing

Struct sizes in C

Here’s an interesting little tidbit about structs and C: the size of the struct depends the order in which you declare your struct’s member variables. This ancient art is called “struct packing”. Let’s look at an example:

#include <stdio.h> 

struct X { int a; char b; int c; }; 

int main(void) { 
  /* note: %zu was added in C99 as a format specifier for size_t type */ 
  printf("%zu\n", sizeof(int)); // 4 
  printf("%zu\n", sizeof(char)); // 1 
  printf("%zu\n", sizeof(struct X)); // 12 
  return 0; 
}

This code makes sense - the size of two ints is 8, one char is 9, and the compiler pads our data to a multiple of 4 for efficiency. Fun fact: in gcc, you can use -fpack-struct to get sizeof(struct X) to return 9. However, you probably don’t want to do this unless you know exactly what you’re doing, because although you save memory this way, the code will take much longer to execute. Moving on!

Let’s add another char to our struct. Where shall we put it? Try appending it to the end:

struct X { int a; char b; int c; char d; };

Compile this and run. The size of struct X should now return 16. Why? Well, it turns out that the compiler will see the first three elements as being of size 12 already, and the extra char d adds 1, giving us 13, and 13 becomes 16. Let’s try putting char d immediately after char b:

struct X { int a; char b; char d; int c; };

Compile and run. Whoa! sizeof(struct X) now returns 12! The compiler now sees the first three elements as being of size 8, and adds the int to return 12.

Moral of the story: place your data in memory-size order in your C structs! Other compilers do this automatically, but C/C++ don’t.

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