Generic Swap

[Benjamin Zhu]

In article <1990Jun11.214844.21531 at agate.berkeley.edu> dankg at sandstorm.Berkeley.EDU (Dan KoGai) writes:
>>swap the actual objects.  But you can't, because this routine doesn't know the>...You could use
>>	void generic_swap(void *pa, void *pb, size_t n) {
>>	    void *ptmp = malloc(n);  assert(ptmp != NULL);
>>	    memcpy(ptmp, pb, n);  memcpy(pb, pa, n);  memcpy(pa, ptmp, n);
>>	}
>>but this is pretty silly.  Just code the bloody thing inline, rather than
>>trying to use a generic routine.
>
>	I agree:  Usually we don't swap objects so large that requires
>memcpy().  Instead we swap pointers to the objects.  But C sometimes
>allows assignment of objects larger than size of register, such as
>structs and some compilers allow to pass structs as arguments (not ptr
>to structs).  Here's my quick & dirty macro
>
>void *temp;
>#define swap(type, x, y) temp = (type *)malloc(sizeof(type));\
>	*(type)temp = x; x = y ; y = *(type)temp; free(temp)
	  ^^^^			       ^^^^
	I suppose you want to write (type *) instead. Otherwise, it
	looks OK. 

>
>	Since this is a macro, it should work even though your compiler
>does not allow passing structs as arguments (struct assignment is valid
>since K&R while struct as argument is not).  But since it's a macro,
>handle with care (even better for C++ #inline)
>
>>[2] Note to readers: please don't followup this thread unless you've already
>>read the relevant part of the Frequently Asked Questions document, and have
>>something *new* to say.  I don't want to see the XOR hack being discussed to
>>death again.
>
>	Since I haven't seen my macro version, I decided to follow.
>
>Dan Kogai (dankg at ocf.bekeley.edu)

Benjamin Zhu

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