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The maximum number that can be correctly processed is 2147483647, which is a 31-bit binary number. An error occurs at the 32nd bit. The version of moderator 3742668 is the same. Entering 2147483648 causes an error. Although the precision can reach 4294967295 by judging the number of shifts, I don't encounter such a large number in application, so I won't bother much.

@echo off

setlocal enabledelayedexpansion

set hexstr=0 1 2 3 4 5 6 7 8 9 A B C D E F

set d=0

for %%i in (%hexstr%) do (set d!d!=%%i&set/a d+=1)

set/p scanf=The maximum number that can be correctly processed is , please enter the decimal number to be converted:

if not defined scanf exit/b

set dec=%scanf%

call :d2h

if not defined hex set hex=0

echo The hexadecimal of %dec% is: 0x%hex%

pause>nul

exit/b



:d2h

if %scanf% equ 0 exit/b

set/a tscanf=%scanf%"&"15

set/a scanf">>="4

set hex=!d%tscanf%!!hex!

goto :d2h

==Deleted==

[ Last edited by 0401 on 2007-1-25 at 09:19 AM ]

Originally posted by electronixtar at 2007-1-7 08:46:
The building owner is a strong person

I'm a noob, and I'm shy.
Post another decimal to binary conversion using the same method.

@echo off

setlocal enabledelayedexpansion

set/p scanf=The maximum number that can be correctly processed is , please enter the decimal number to be converted:

if not defined scanf exit/b

set dec=%scanf%

call :d2b

echo The binary of %dec% is: %bin%

pause>nul

exit/b



:d2b

set a=1

for /l %%i in (1,1,31) do (

	set/a t=%scanf%"&"!a!

	set/a a*=2

	if !t! equ 0 (set bin=0!bin!) else (set bin=1!bin!)

)

if %bin% equ 0 set bin=0 & goto :eof

:cut0

if %bin:~0,1% equ 0 set bin=%bin:~1%&goto :cut0

Super top! Appreciate ~~ : )

It's really refreshing to use, amazing.

That's really amazing. I'm supporting it.

Brother 0401 on the 4th floor, the decimal to binary conversion can still be improved. The tag :cut0 at the end is redundant. It's better to be like this:

:d2b
set a=1
for /l %%i in (1,1,31) do (
set/a t=%scanf%"&"!a!
set/a a*=2
if !t! equ 0 (set bin=0!bin!) else (set bin=1!bin!)
if %scanf% lss !a! exit/b
)
if %bin% equ 0 set bin=0 & goto :eof

Modifications are as follows:
1: 1,1,30 is changed because for /l %%i in (1,1,31) do (set/a a*=a) will already have an error.
2: Added a line if %scanf% lss !a! exit/b, because the operation of number a and a larger number is always 0. Since this is the case, it's better to directly end with if %scanf% lss !a! exit/b, and then the preceding 0s won't be removed later.

Add a "stamp" first, otherwise it won't be found next time!

I will reply to post 9

if %scanf% lss !a! exit/b
There are obvious drawbacks in your code.
Every time the loop is executed, a judgment is made, which reduces the program efficiency

There is another question
"1,1,30 Because you for /l %%i in (1,1,31) do (set/a a*=a) has already gone wrong"

When %%i = 31 here, a = 2^(30), there is no mistake. It will be wrong when %%i = 32, at this time a = 2^(31)

So the code on floor 9 is correct!

I used the method of remainder to write a decimal to hexadecimal conversion, and confirmed that the maximum limit processing is indeed 2147483647, that is, the hexadecimal limit is 7FFFFFFF, and the reversal also illustrates this problem. It should be said to be the limit of CMD.