Alright, here you go. Why doesn't this code work?
No matter which number you enter, the program will
always treat it as the number 5. Why?

#include <stdio.h>
#include <math.h>

int main()
{
  int x;
  
  printf("Enter any value but 5: ");
  scanf("%d", &x);
  getchar();

  x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;

  if(x=!5)  
    printf("You didn't type '5'! Congratulations.\n");
  else
    printf("Bleh, you typed '5'.\n");
    
  return 0;
}

:   x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;

Hmm, if I'm not wrong, shouldn't there be a = after the >> or something?

And it seems to alwasy crash for me... Does it need to be compiled as a dos project or something to run? or is it just my compiler (could be becouse it's bad).

:

: :   x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;
: 

:
: Hmm, if I'm not wrong, shouldn't there be a = after the >> or something?
:
: And it seems to alwasy crash for me... Does it need to be compiled as a dos project or something to run? or is it just my compiler (could be becouse it's bad).
:


It crashes in MS-Windows 32-bit compiler because its attempting to access memory it does not own.

: :

: : :   x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;
: : 

: :
: : Hmm, if I'm not wrong, shouldn't there be a = after the >> or something?
: :
: : And it seems to alwasy crash for me... Does it need to be compiled as a dos project or something to run? or is it just my compiler (could be becouse it's bad).
: :
:
:
: It crashes in MS-Windows 32-bit compiler because its attempting to access memory it does not own.
:


Yeah probably. But from a platform-independant point of view, it shouldn't. The solution is actually very simple, you don't even need to run it to find it. And nope, there shouldn't be a = after >>

:
: Yeah probably. But from a platform-independant point of view, it shouldn't. The solution is actually very simple, you don't even need to run it to find it. And nope, there shouldn't be a = after >>
:

OK, I was a little confused there, becouse it looked very wrong...
I fixed your typing errors, but they where the sollution....

With some equation solving:

x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;
x >> (*(int*)(int)(.76) ^ 0xFF00 % 33) % 8;
x >> (*(int*)1 ^ 0xFF00 % 33) % 8;

Then it needs to get what comes after NULL, which should make every program crash. How does it run on every platform then? (if the compiler doesn't optimise away the statement)

: :
: : Yeah probably. But from a platform-independant point of view, it shouldn't. The solution is actually very simple, you don't even need to run it to find it. And nope, there shouldn't be a = after >>
: :
:
: OK, I was a little confused there, becouse it looked very wrong...
: I fixed your typing errors, but they where the sollution....
:
: With some equation solving:
:

: x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;
: x >> (*(int*)(int)(.76) ^ 0xFF00 % 33) % 8;
: x >> (*(int*)1 ^ 0xFF00 % 33) % 8;
: 

:
: Then it needs to get what comes after NULL, which should make every program crash. How does it run on every platform then? (if the compiler doesn't optimise away the statement)
:


Heh, you are going to be pissed off when I tell you the solution.
It is a tricky question: I already gave the hint that x >> was not a typo...

: : :
: : : Yeah probably. But from a platform-independant point of view, it shouldn't. The solution is actually very simple, you don't even need to run it to find it. And nope, there shouldn't be a = after >>
: : :
: :
: : OK, I was a little confused there, becouse it looked very wrong...
: : I fixed your typing errors, but they where the sollution....
: :
: : With some equation solving:
: :

: : x >> (*(int*)(int)(43.54*sin(3.14/180.0)) ^ 0xFF00 % 33) % 8;
: : x >> (*(int*)(int)(.76) ^ 0xFF00 % 33) % 8;
: : x >> (*(int*)1 ^ 0xFF00 % 33) % 8;
: : 

: :
: : Then it needs to get what comes after NULL, which should make every program crash. How does it run on every platform then? (if the compiler doesn't optimise away the statement)
: :
:
:
: Heh, you are going to be pissed off when I tell you the solution.
: It is a tricky question: I already gave the hint that x >> was not a typo...
:

I've already found the sullotion, but it still crashes.
(*(int*)1) will be segmentation fault on all unixes, and the code is therefor NOT platformindependent.

:
: I've already found the sullotion, but it still crashes.
: (*(int*)1) will be segmentation fault on all unixes, and the code is therefor NOT platformindependent.
:

I agree -- very similar to this code, which will also crash

int x = *(int *)0;

: :
: : I've already found the sullotion, but it still crashes.
: : (*(int*)1) will be segmentation fault on all unixes, and the code is therefor NOT platformindependent.
: :
:
: I agree -- very similar to this code, which will also crash
:

: int x = *(int *)0;
: 

:
:


Alright, I agree, you found one problem but not the solution.

No matter which number you enter, the program will
always treat it as the number 5. Why?

: : :
: : : I've already found the sullotion, but it still crashes.
: : : (*(int*)1) will be segmentation fault on all unixes, and the code is therefor NOT platformindependent.
: : :
: :
: : I agree -- very similar to this code, which will also crash
: :

: : int x = *(int *)0;
: : 

: :
: :
:
:
: Alright, I agree, you found one problem but not the solution.
:
: No matter which number you enter, the program will
: always treat it as the number 5. Why?
:

I didn't want to give away the solution to the OP.

But anyway here it is:

  if(x=!5)  
    printf("You didn't type '5'! Congratulations.\n");

The condition will never return true, becouse !5 is 0 and x=0 is 0, which is false.