Why can std:set (with a single colon) compile?
I accidentally wrote
std::set<string> keys;
as:
std:set<string> keys;
but weirdly enough, Visual Studio 2013 still compiles.
Why does this happen?
Actually keys is not only defined, but later used as a set of strings, such as
if(keys.find(keystr)==keys.end()){
keys.insert(keystr);
toret.push_back(tempv);
}
At block scope, an identifier followed by a single colon introduces a label. Thus, your statement is equivalent to:
set<string> keys;
except that it bears the label std and can be jumped to by the statement goto std;.
For some reason, the name set is known to the compiler---perhaps you did using namespace std;, or using std::set;, or something like that, or perhaps you defined your own set type somewhere.
In the second case, std is a label. It is the same as spelling default incorrectly in a case statement.
std::set<string> keys;
as:
std:set<string> keys;
but weirdly enough, Visual Studio 2013 still compiles.
Why does this happen?
Actually keys is not only defined, but later used as a set of strings, such as
if(keys.find(keystr)==keys.end()){
keys.insert(keystr);
toret.push_back(tempv);
}
At block scope, an identifier followed by a single colon introduces a label. Thus, your statement is equivalent to:
set<string> keys;
except that it bears the label std and can be jumped to by the statement goto std;.
For some reason, the name set is known to the compiler---perhaps you did using namespace std;, or using std::set;, or something like that, or perhaps you defined your own set type somewhere.
In the second case, std is a label. It is the same as spelling default incorrectly in a case statement.
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