Consider the following line of code:
new (p++) T();
If the constructor T() throws an exception, is p guaranteed to have already been incremented?
Consider the following line of code: new (p++) T(); If the constructor T() throws
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Consider the following line of code:
new (p++) T();
If the constructor T() throws an exception, is p guaranteed to have already been incremented?
From 5.3.4 [expr.new] (quoting from n3242):
So in a new expression the allocation function is used from a function call (which makes sense). All allocation functions are functions, including the ones provided by the implementation, from 3.7.4.1 [basic.stc.dynamic.allocation]:
So by the time an exception is thrown from the constructor, the allocation has taken place and the associated function call expression has been fully evaluated.