On 08/12/2016 07:43, topher wrote:
> #include <string>
> #include <cstdio>
> #include <sstream>
>
> using namespace std;
>
> struct StaticName {
> string getName() {
> static int i = 0;
> name.clear();
> ostringstream oss;
> oss << ++i;
> name += oss.str();
> return name;
> }
> static string name;
> };
> string StaticName::name = "";
>
> int main(int argc, char* argv[]) {
> StaticName sn;
> printf("name1=%s, name2=%s, name3=%s\n",
> sn.getName().c_str(), sn.getName().c_str(), sn.getName().c_str());
> return 0;
> }
>
> The executable built by g++ 4.8.5 prints "name1=3, name2=2, name3=1".
> However, executable built by clang++ 3.8.0 prints "name1=1, name2=2,
> name3=3", which is what I expected.
> Is there any undefined behavior in my code?
I can't answer for C++ but in C99:
6.5.2.2 Function calls
> 10) The order of evaluation of the function designator, the actual
> arguments, and subexpressions within the actual arguments is
> unspecified, but there is a sequence point before the actual call.
So basically, foo(f1(), f2(), f3()); might evaluate in any order:
f1, f2, f3
f1, f3, f2
f2, f1, f3
f2, f3, f1
f3, f1, f2
f3, f2, f1
I suppose the same rule applies to C++
And note that unspecified behavior is slightly different than
undefined behavior.
> 1 unspecified behavior
> use of an unspecified value, or other behavior where this
> International Standard provides two or more possibilities and imposes
> no further requirements on which is chosen in any instance
> 2 EXAMPLE
> An example of unspecified behavior is the order in which the
> arguments to a function are evaluated.
Regards.
