Question

I have two data structures:

typedef struct{
    int a;
    int b;
    int c;
}EVENTS;

EVENTS typeone[20];
EVENTS typetwo[20];

These have been filled. typeone has been filled till typeone[5] and typetwo till typetwo[8].

I just want to compare the first six of typeone and typetwo and see if there are equal in all their members.

Is there a way to do typeone[1] == typetwo[1]
Basically comparing all the values inside the datastructure at [1].
Is there a short way to do this or would I have to loop through each member and compare separately?

Thanks

Answer 1

 typedef struct{
     int a;
     int b;
     int c;
 }EVENTS;

 #pragma pack(1)
     EVENTS typeone[20];
     EVENTS typetwo[20];
 #pragma pack()

 int equal(EVENTS* v1, EVENTS* v2)
 {
      return 0==memcmp(v1, v2, sizeof(*v1));
 }

Note #pragma pack(1). It ensures that there are no padding bytes in the structures. This way memcmp will not try to compare padding bytes and the comparison is way faster than a field-by-field method, but while in this case the performance is unlikely to be adversely affected, take:

     typedef struct{
         char a;
         long b;
     } somestruct;

 #pragma pack(1)
     somestruct foo;
 #pragma pack()

Retrieving foo.b will take much more machine code than in case of padded structures, because it will miss word-aligned position where it can be retrieved with a single 32-bit instruction, it will have to be picked out with four byte-reads, and then assembled into the target register from these four pieces. So, take the performance impact into account.

Also, check if your compiler supports #pragma pack. Most modern compilers do, but exceptions may still happen.