c - Why is the compiler confused into a SIGSEGV by unsigned int? -

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I have solved a problem, which has given me a sense of what is going on, but still not at all. 

  int main () {unsigned int a = 2; Four c [2] = {}; Char * p = & amp; C [1]; Return P [1 - A]; }   
 This is a bit clearer when the last line is rewritten.  
  return * (P + (1 - A)); / * Equivalent * / return * (P + 1 - A); / * Work * / return * (P + (1 - (int))); / * Functions * /   
 I'm surprised that the compiler does not remove the bracket internally. And more so that it clearly tries to capture a temporary negative result of type  unsigned int  unless there is a reason for division fault. The output of the assembler has a slight difference between the brackets and without the code.  
  - movl $ 1,% eax - subl-12 (% rbp),% eax - movl% eax,% edx + movl-12 (% rbp),% eax + movl $ 1 ,% edx + subq% rax,% rdx    
 
 
 Regarding this forcible rules The expression is  1-a  as a  unsigned int , and the result is in an underflow. The compiler can not remove the bracket because you are mixing the types, consider your cases:  
  return * (P + (1 - A)); / * Equals * /   
 first calculates  1-a , but treats it as a  unsigned int . This is less than an unsigned type, and gives maximum value for  unsigned int  it is then added to the indicator, resulting in  p + (1 <31) , if  unsigned int  is 32-bit, such as an indicator is removed for some, it is unlikely to be a valid memory location.  
  return * (P + 1 - A); It calculates  p + 1  and then decreases  a , resulting in dereferencing  p- 1 . This is technically undefined behavior, but perhaps (in most implementations) a reference to a valid memory spot on the stack.  
  return * (p + (1 - (int) a)); / * Functions * /   
 This adds a  a  to a signed  int , and then  1-A < / Counts the code, which is  -1 . You can again dereference  p-1 .

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