CODEWITHSUNDEEP
c++functionsegmentation-fault
user2304458
user2304458

Reputation: 387

No Segmentation fault why? returned function value used as parameter

Can the code below cause segmentation fault?

struct Dim {
  int x, y, z;
};

Dim set_dim(int a) {
 Dim l;
 l.x=a;
 l.y=a;
 l.z=a;
 return l;
}

int sum(const Dim &m) {
  int s=m.x+m.y;
  return s;
}

main() {
  cout<<sum(set_dim(5))<<endl;
}

I think it can because a reference is taken of a local variable 'l' in set_dim, in other word a reference to variable now out of scope. But never the less it works

Upvotes: 1

Views: 145

Answers (3)

stardust
stardust

Reputation: 5998

A reference of l is never taken (not directly may be due to return value optimizations indirectly but you don't have to worry about that).

  • set_dim returns a copy of l. A temporary copy which will be passed to sum.

  • sum can accept a temporary because it takes a const reference and everything worked out just fine and they all lived happily until sum returned.

Upvotes: 3

lthms
lthms

Reputation: 519

set_dim does not return a reference but a copy of l, so I don't think there is any chance of seg fault.

Upvotes: 1

Joseph Mansfield
Joseph Mansfield

Reputation: 110748

This is well-defined. set_dim returns by value, so its returned object is a temporary copy of the local variable l. This temporary is then bound to the const reference argument of sum. This affects the lifetime of the temporary object:

A temporary bound to a reference parameter in a function call (5.2.2) persists until the completion of the full-expression containing the call.

So the lifetime of the temporary object is the full-expression cout<<sum(set_dim(5))<<endl;. It still exists while sum is executing.

Upvotes: 3

Related Questions