Return by link

Question

See the following code snippets. In the second function, I return the link. I declare a local variable in the function and return the address. Since the variable is local, I believe that her life ends when she leaves the function. My question is, why is it possible to access the value from the caller without any exception, even if the original variable is deleted?

int& b=funcMulRef(20,3);

int* a= funcMul(20,3);


int* funcMul(int x,int y)
{
 int* MulRes = new int;
      *MulRes = (x*y);

 return MulRes;

}

int& funcMulRef(int x,int y)
{
 int MulRes ;
      MulRes = (x*y);

 return MulRes;

}

Regards, JOHN

Answer 1

The behavior of the second function is simply undefined; anything can happen, and in many cases it will work, simply because nothing is overwritten where the result was used to be stored on the stack.

Answer 2

, .

, , , , , , , .

, - , , . .

Answer 3

. , .

, undefined.

, .

Answer 4

, . undefined. , , .

Are you trying to avoid temporary objects? If so, you might be interested:
http://en.wikipedia.org/wiki/Return_value_optimization

Answer 5

Most likely, this will not work in these cases:

funcMulRef(10,3) + funcMulRef(100,500)

alternatively, in a more unpleasant way:

std::cout << "10*3=" << funcMulRef(10,3) << " 100*500=" << funcMulRef(100,500) << std::endl;

gcc will warn of errors of this type if you use -Wall