[Solved] Can't I use CRTDLL.DLL in Win64?

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A$M

Joined: 29 Feb 2012
Posts: 94

A$M 05 Feb 2014, 19:07

Hi,

I'm trying to use AVX. AVX is not the problem. "sprintf" is. Can't I use it?
My code:

Code:

format PE64 CONSOLE
entry main
include 'win64ax.inc'

section '.text' code readable executable

main:
  sub rsp, 8
  vsqrtpd ymm0, qqword[value]
  vmovaps qqword[result], ymm0
  cinvoke sprintf, msg, string, [value], [value+8], [value+16], [value+24]
  invoke MessageBoxA, 0, msg, 0, MB_OK
  invoke ExitProcess, 0

section '.data' data readable writeable
  msg rb 64
  string db "1.0 -> %f, 2.0 -> %f, 3.0 -> %f, 4.0 -> %f,",0
  align 32
  value dq 1.0
        dq 2.0
        dq 3.0
        dq 4.0

section '.bss' data readable writeable
  result rq 4

section '.idata' import data readable writeable

 library kernel32,'KERNEL32.DLL',\
         user32,'USER32.DLL',\
         crtdll,'CRTDLL.DLL'

 include "API\KERNEL32.INC"

 import user32,\
        MessageBoxA,'MessageBoxA'

 import crtdll,\
        sprintf,'sprintf'

Ps.: "Can't I use" is correct in English?

Last edited by A$M on 05 Feb 2014, 19:58; edited 1 time in total

05 Feb 2014, 19:07

tthsqe

Joined: 20 May 2009
Posts: 767

tthsqe 05 Feb 2014, 19:21

work for me with 'MSVCRT.DLL' in place of 'CRTDLL.DLL'.
Why don't you write your own floating point print functions?

05 Feb 2014, 19:21

HaHaAnonymous

Joined: 02 Dec 2012
Posts: 1178
Location: Unknown

HaHaAnonymous 05 Feb 2014, 19:31

[ Post removed by author. ]

Last edited by HaHaAnonymous on 28 Feb 2015, 18:18; edited 1 time in total

05 Feb 2014, 19:31

A$M

Joined: 29 Feb 2012
Posts: 94

A$M 05 Feb 2014, 19:58

tthsqe wrote:

work for me with 'MSVCRT.DLL' in place of 'CRTDLL.DLL'.

tthsqe, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks, thanks. Very Happy

tthsqe wrote:

Why don't you write your own floating point print functions?

I don't need if I have MSVCRT and CRTDLL. Wink

HaHaAnonymous wrote:

Quote:

Ps.: "Can't I use" is correct in English?

Yes, it is.

Thanks you too, HaHaAnonymous. My English is not so bad. Smile

05 Feb 2014, 19:58

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