std::map operator[] and default values of build-in types

[qfel]

If new pair is inserted into std::map through operator[], is default value
for build-in types equal to 0?
eg.
std::map<std::string,bool> m;
bool b=m["not existing key"];
will b equal to false, or will it contain undefined value?

 

[David Harmon]

On Tue, 29 Mar 2005 11:15:22 +0200 in comp.lang.c++, "qfel"

If new pair is inserted into std::map through operator[], is default value
for build-in types equal to 0?
Yes.

eg.
std::map<std::string,bool> m;
bool b=m["not existing key"];
will b equal to false, or will it contain undefined value?

The std containers go to some effort to avoid undefined values.

 

[Howard Hinnant]

If new pair is inserted into std::map through operator[], is default value
for build-in types equal to 0?
eg.
std::map<std::string,bool> m;
bool b=m["not existing key"];
will b equal to false, or will it contain undefined value?

b will be false.

-Howard