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Original post by i1977 Washu, the reason I can't use memset is that even though it takes an int as a parameter, it only copies 1 byte repeatedly, not 4 bytes like what I need to do.

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Original post by Dmytry also,i'd compare performance of rep with performance of loop.... rep might be actually slower :( .

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From How to Optimize for the Pentium® Family of Microprocessors (pentopt.pdf) section 18.4 ... REP MOVSD and REP STOSD are quite fast if the repeat count is not too small. Always use the DWORD version if possible, and make sure that both source and destination are aligned by 8. ... On PPro, P2 and P3, REP MOVS and REP STOS can perform fast by moving an entire cache line at a time. This happens only when the following conditions are met: • both source and destination must be aligned by 8 • direction must be forward (direction flag cleared) • the count (ECX) must be greater than or equal to 64 • the difference between EDI and ESI must be numerically greater than or equal to 32 • the memory type for both source and destination must be either write-back or writecombining (you can normally assume this). Under these conditions, the number of uops issued is approximately 215+2*ECX for REP MOVSD and 185+1.5*ECX for REP STOSD, giving a speed of approximately 5 bytes per clock cycle for both instructions, which is almost 3 times as fast as when the above conditions are not met.

DWORD Call_cdecl( const void* args, size_t sz, DWORD func )
{
    DWORD rc;               // here's our return value...
    __asm
    {
        mov   ecx, sz       // get size of buffer
        mov   esi, args     // get buffer
        sub   esp, ecx      // allocate stack space
        mov   edi, esp      // start of destination stack frame
        shr   ecx, 2        // make it dwords
        rep   movsd         // copy params to real stack
        call  [func]        // call the function
        mov   rc,  eax      // save the return value
        add   esp, sz       // restore the stack pointer
    }
    return ( rc );
}

void* memset4(void* t, unsigned int val, size_t count) {
	void *dst = t;
	while(count--) {
		*(unsigned int*)dst = val;
		dst = (unsigned int*)dst + 1;
	}
	return t;
}

004014BD B8 34 12 00 00   mov         eax,1234h 
004014C2 B9 64 00 00 00   mov         ecx,64h 
004014C7 8D 7C 24 08      lea         edi,[esp+8] 
004014CB F3 AB            rep stos    dword ptr [edi] 

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Original post by i1977 I will use inline assembly code anyway because the code generated for the Debug build target is much less optimized.

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Original post by i1977 Washu, When I said "compiler specific", I wasn't refering to Microsoft extensions such as __asm. I know that's not portable and I don't really care since I won't be compiling my code with another compiler anyway. What I meant is that your C function might not compile to the same optimized asm code on another compiler such as gcc, Borland or whatever. I tried compiling your C function and it does indeed produce optimized assembly code for the Release build target. I must admit that I am very surprised about this, but nonetheless, I will use inline assembly code anyway because the code generated for the Debug build target is much less optimized. Oh, and what does "HRM????" mean by the way?

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Original post by CGameProgrammer I believe memset is "rep stosb" and memcpy is "rep movsb". In other words they are both single-byte only (what the last 'b' stands for).

CopyUp:
        test    edi,11b         ;U - destination dword aligned?
        jnz     short CopyLeadUp ;V - if we are not dword aligned already, align

        shr     ecx,2           ;U - shift down to dword count
        and     edx,11b         ;V - trailing byte count

        cmp     ecx,8           ;U - test if small enough for unwind copy
        jb      short CopyUnwindUp ;V - if so, then jump

        rep     movsd           ;N - move all of our dwords

        jmp     dword ptr TrailUpVec[edx*4] ;N - process trailing bytes

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Original post by i1977 Thanks for your help guys! Even Washu, who apparently knows more about what I need and am trying to do. ;)