Geek asks and answers

Can Lithuanian int assignment in C ++ throw an exception?

This code is as follows:

void f()
{
   int i;
   i = 0;
}

Is it possible that the system may throw an exception due to a simple assignment?

[Edit: For those who say, “No exception can happen,” can you point Me toward the part of the C ++ standard that talks about this? It's hard for me to find him.]

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5 answers

Although it will probably be difficult for you to find confidence in it in the standard, a simple rule of thumb is that anything that is legitimate in C may not give up. [Edit: the closest that I know of, a direct statement on this subject is in §15 / 2, which states that:

, throw, "throw exception"; [...]

, , throw, .]

: UB. - ++, , , operator = new.

. , , . , , . , , - .

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, , :

int operator"" _t(const char *) { throw 0; } // C++11 user defined literal 

struct foo {
  foo(int) { throw 0; }
  operator int() { throw 0; }
  foo& operator=(int) { throw 0; }
};

int main() {
  int i;
  i = 0; // can't throw
  i = 0_t; // User defined literal throws
  foo f = 0; // Constructor throws
  i = f; // conversion operator throws
  f = 0; // assignment throws
  f = f; // both conversion and assignment would like to throw
}

( ++ 11)

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, 0 ( int) int, § 5.17 , , . int, §5.17 , " cv-unqualified . " :

  • , int, ( C : int , , )

  • , undefined, int, undefined ( )

  • , . .

: , , , . , operator=. , , .

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: operator=() ; , , operator T() . , - , , , .

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int, - .

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