goto crosses initialization [duplicate]

Question

Let's say I have some c++ code:

if (error)
    goto exit;
... 
// size_t i = 0; //error
size_t i;
i = 0;
...
exit:
    ...

I understand we should not use goto, but still why does

size_t i;
i = 0;

compile whereas size_t i = 0; doesn't?

Why is such behavior enforced by the standard (mentioned by @SingerOfTheFall)?

Answer 1

You can't jump over initialization of an object.

size_t i = 0; 

is an initialization, while

size_t i;
i = 0;

is not.

The C++ Standard says:

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps from a point where a local variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has POD type (3.9) and is declared without an initializer.