The lifetime-extension of a temporary object can be performed only once, when the temporary object gets bound to the first reference. After that, the knowledge that the reference refers to a temporary object is gone, so further lifetime extensions are not possible. The case that is puzzling you A const& refnop = A(a).nothing(); is similar … Read more
There is no difference unless you write something weird like (v1 += v2) = v3; In the first case, the assignment will be to a temporary, and the overall effect will be v1 += v2. In the second case, the assignment will be to v1, so the overall effect will be v1 = v3. In … Read more
1. Is getA1() implementation correct ? I feel it is incorrect as it is returning address of local variable or temporary. The only version of getAx() that is correct in your program is getA3(). Both of the others have undefined behaviour no matter how you use them later. 2. Which of the statements in main … Read more
Only local const references prolong the lifespan. The standard specifies such behavior in ยง8.5.3/5, [dcl.init.ref], the section on initializers of reference declarations. The reference in your example is bound to the constructor’s argument n, and becomes invalid when the object n is bound to goes out of scope. The lifetime extension is not transitive through … Read more