Re: Can a well-formed program incur undefined behaviour?
jordigh@gmail.com wrote:
Reading 1.3.12 of the current standard and comparing it with 1.3.13,
it's not clear to me if the standard allows a well-formed program
(1.3.14) to incur undefined behaviour. In Freenode's ##c++ channel,
people suggested that
#include<vector>
int main(){
std::vector<int> x; x[2] = 5;
}
is a well-formed program that causes undefined behaviour, but doesn't
this contradict the requirement of a well-formed program that it is
"constructed according to [...] diagnosable semantic rules"?
The definition of "diagnosable rule" is given in 1.4p1. It does not include
undefined behavior and rules for which it is stated that "no diagnostic is
required".
The above is well-formed only if we don't violate a diagnosable rule. Table
68 defines op[](N) to *(.begin() + N). In your case, .begin() is not
dereferencable, so you can do x[N] only for N==0. Since you don't, you
violate the precondition stated in Table 74 for "++r".
I can't find where the spec says what happens if a precondition of a
Requirement expression is violated, but I think behavior is undefined, which
would mean you don't violate a diagnosable rule.
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