I$$d like to ask an expert.

#include

#include

#include

$//$#include $"../../$shared$/$kernels$\_$lib.c$"//$Uncomment this if kernels$\_$lib.c is required

char name$[12]$;

void call$\_$me$()\{$

printf$("$call$\_$me$()$function was called!

$")$;

$\}$

void unsafe$()\{$

char buffer$[420]$;

int characters$\_$read;

u$\_$int$32\_$t XORbius $=0$; $//$Ensure XORbius is initialized to $0$

u$\_$int$32\_$t XOR$\_$string $=0$xdeadbeef;

printf$("$Feed Me A Stray String:

$")$;

characters$\_$read $=$read$(0,$buffer$,420)$;

$//$Loop to construct XORbius byte$-$by$-$byte with debug output

for $($int i $=0$; i $<4$; i$++)\{$

XORbius $=$XORbius $<<8$; $//$Shift XORbius by $8$bits for the next byte

u$\_$int$8\_$t buffer$\_$byte $=($unsigned char$)$buffer$[201+($i $*2)]$;

u$\_$int$32\_$t xor$\_$byte $=($XOR$\_$string $>>($i $*8))$& $0$xFF;

XORbius $|=($buffer$\_$byte $^$xor$\_$byte$)$;

$//$Debug statements to observe values at each step

printf$("$Iteration $\%$d:

$",$i$)$;

printf$("$buffer$[\%$d$]=0$x$\%02$x

$",201+($i $*2),$buffer$\_$byte$)$;

printf$("$XOR$\_$string byte $=0$x$\%02$x

$",$xor$\_$byte$)$;

printf$("$Resulting XORbius $=0$x$\%08$x

$",$XORbius$)$;

$\}$

$//$Final check

if $($XORbius $==0$xdeadbeef$)\{$

call$\_$me$()$;

$\}$else $\{$

printf$("$XORbius did not match $0$xdeadbeef, it was $0$x$\%08$x

$",$XORbius$)$;

$\}$

$\}$

int main$($int argc, char$*$argv$[])\{$

char $*$bin$=($char $*)$malloc$(5)$;

strcpy$($bin$,$&argv$[0][$strlen$($argv$[0])-4])$;

if $($strcmp$($bin$,"$flag$")!=0)\{$

printf$("$ERROR DETECTED: MODIFIED BINARY NAME $\%$s

$",$bin$)$;

exit$(-1)$;

$\}$

unsafe$()$;

return $0$;

$\}$

You need to unravel the logic that this program is checking against in order to get to the call$\_$me$()$function$!,$simply$$need to input the right values that will correctly decode the logic and pass the checks.

XOR TRUTH TABLE

A B Result

$000$

$011$

$101$

$110$

XOR Operations are REVERSIBLE, meaning that performing the same XOR operation on a number twice will end up with the original number!

Example:

$0$b$11110000^0$b$00001111=0$b$11111111$

$0$b$11111111^0$b$00001111=0$b$11110000$

in Hex notation

$0$xF$0^0$x$0$F $=0$xFF

$0$xFF $^0$x$0$F $=0$xF$0$

I tried this payload in my python file : payload $=$b$"$A$"*201+$b$"\backslash$x$31"+$b$"$A$"+$b$"\backslash$x$13"+$b$"$A$"+$b$"\backslash$x$73"+$b$"$A$"+$b$"\backslash$x$31"+$b$"$A$"*(420-209)$

Iteration $0$: buffer$[201]=0$x$31$XOR$\_$string byte $=0$xef Resulting XORbius $=0$x$000000$de

Iteration $1$: buffer$[203]=0$x$13$XOR$\_$string byte $=0$xbe Resulting XORbius $=0$x$0000$dead

Iteration $2$: buffer$[205]=0$x$73$XOR$\_$string byte $=0$xad Resulting XORbius $=0$x$00$deadde

Iteration $3$: buffer$[207]=0$x$31$XOR$\_$string byte $=0$xde Resulting XORbius $=0$xdeaddeef XORbius did not match $0$xdeadbeef, it was $0$xdeaddeef

Please don't use the solution that is already here in Chegg it didn't work and don't use AI$,$I have already tried.