What happens if you transfer control to a if(false) block by using goto?

Question

I've thought of following code by trying to solve a difficult 'nested-condition' problem:

goto error;
    
if (false)
{
error:
    cout << "error block" << endl;
}
else
{
    cout << "else block" << endl;
}

When I run this code, only error block is displayed, as expected (I guess?). But is this defined behavior across all compilers?

Answer 1

Yes, this is well defined. From stmt.goto#1

The goto statement unconditionally transfers control to the statement labeled by the identifier. The identifier shall be a label located in the current function.

There are some restrictions, e.g. a case label cannot cross a non-trivial initialization

goto error;
int i = 42;
error:       // error: crosses initialization of i

But these don't apply to your example. Also, in the case of crossing an initialization, this is a hard compiler error, so you don't have to worry about undefined behavior.

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Note that once you jump to the case label error, you're effectively inside the true branch of the if condition, and it doesn't matter that you got there via a goto. So you're guaranteed that the else branch will not be executed.