Say i have:
unsigned char *varA, *varB, *varC; varA=malloc(64); varB=malloc(32); varC=malloc(32); How can i put the first 32 byte of varA into varB and the last 32 byte of varA into varC?
- Your question has remarkable little to do with pointers and you might want to rethink what it means to "split them".pmr– pmr2013-01-30 22:37:58 +00:00Commented Jan 30, 2013 at 22:37
- 2I would recommend a union... just to clarify, i'm suspecting it's gonna be bits instead of bytes in the end.Shark– Shark2013-01-30 22:37:59 +00:00Commented Jan 30, 2013 at 22:37
Add a comment |
3 Answers
memcpy(varB, varA, 32); memcpy(varC, varA + 32, 32); It's this simple because the underlying data type is unsigned char, which is the same size as a byte. If varA, varB, and varC were integers, you would need to multiply the size parameter to memcpy (i.e. 32) by sizeof(int) to compute the right number of bytes to copy. If I were being pedantic, i could have multiplied 32 by sizeof(unsigned char) in the example above, but it is not necessary because sizeof(unsigned char) == 1.
Note that I don't need to multiply the 32 in varA + 32 by anything because the compiler does that for me when adding constant offsets to pointers.
One more thing: if you want to be fast, it might be sufficient to just work on each half of varA separately, rather than allocate two new buffers and copy into them.
2 Comments
polslinux
First of all thanks a lot for your answer, you have been very clear and exhaustive :) Then how can i work with "half of varA"? Let's say i need to do
void *function(int, half_of_varA);, how can i pass only "half of memory"?
Randall Cook
You would choose one of
void* p = function(1234, varA); or void* p = function(1234, varA + 32); depending on which half of varA you need to use. And I am using 1234 as a guess as to the first parameter of function(), of course.You could use loop to copy individual bytes one by one:
for (int i = 0; i != 32; ++i) varB[i] = varA[i]; for (int i = 0; i != 32; ++i) varC[i] = varA[32 + i]; Or memcpy function from the C runtime library:
memcpy(varB, varA, 32); memcpy(varC, varA + 32, 32); 
3 Comments
Randall Cook
Loops certainly work, but I would trust the compiler and/or
memcpy implementer in this case. They can do tricks in the underlying assembly that a compiled loop can't touch (at least without a great optimizer), like moving multiple bytes at once.
arabesc
It's just a different approaches. Copying through the loop could include some processing, like bit shuffling for example.
Randall Cook
Yes, if processing is required, a loop is necessary.
Ok let's do this....
uint64 source; uint32 upperBytes, lowerBytes; upperBytes = source&0xFFFFFFFF00000000; lowerBytes = source&0x00000000FFFFFFFF; Homework done.
1 Comment
polslinux
I don't understand your answer.