While attempting to create a memory manager for future C programs, I've come across this question:
For instance, consider the following struct.
typedef struct { int field1; int field2; char field3; } SomeType; When allocated, will the memory addresses of the fields be in the order field1, field2, field3? Or is this not guaranteed?
Short answer: they are allocated with the order as they declared in the struct.
Example:
#include <stdio.h> #include <string.h> struct student { int id1; int id2; char a; char b; float percentage; }; int main() { int i; struct student record1 = {1, 2, 'A', 'B', 90.5}; printf("size of structure in bytes : %d\n", sizeof(record1)); printf("\nAddress of id1 = %u", &record1.id1 ); printf("\nAddress of id2 = %u", &record1.id2 ); printf("\nAddress of a = %u", &record1.a ); printf("\nAddress of b = %u", &record1.b ); printf("\nAddress of percentage = %u",&record1.percentage); return 0; } Output:
size of structure in bytes : 16 Address of id1 = 675376768 Address of id2 = 675376772 Address of a = 675376776 Address of b = 675376777 Address of percentage = 675376780 The pictorial representation of above structure memory allocation is given below. This diagram will help you to understand the memory allocation concept in C very easily.
Further reading: check out here (also the source for the above example) for C – Structure Padding and Structure dynamic memory allocation in C.
You are guaranteed that field3 comes after field2, which comes after field1, and that field1 is at the start of the memory (i.e. there's no padding before field1). However, they may be padding in between the other members (and even after field3). In short, the order in which they are declared is the order in which they are laid out in memory, though exact alignment and padding is implementation defined (but there won't be any padding before the first member).
SomeType x; ((void*) &x) == ((void*) &x.field1).
field1 and field2, you can say that padding is after field1, or you can alternatively say there is padding before field2. Neither statement is incorrect; it just depends on perspective. But if I misinterpreted your question, and you were instead wondering why there might be any padding at all, then the answer is all about aligning things in memory with the processors native alignment requirements.int is aligned on 4-byte boundaries (if int is 4 bytes), while char has no such restriction (it as a 1-byte alignment, so can be anywhere in memory). If you have an int after a char, 3 bytes of padding are needed to make sure the int lines up on a memory address that's divisible by 4.For example:
typedef struct { char field1; int field2; double field3; } SomeType; The struct's first location is x:
field1's length is 1(byte), filed2 is 4, field3 is 8
so field1 is at x+0, field2 is at x+4~x+7, field3 is at x+8~x+15, and x+1~x+3 is padded to make field2 at right place.
And the whole struct's length should be divide exactly by its largest member; if not, padding some bytes to the end.
