#include <string>
using namespace std;
int main()
{
string s = "bob";
const char *t = s.c_str();
return 0;
}
string to char ptr?
Author
does this even work and also does it have undefiend behavior?
It works, and is what c_str is for. The buffer returned by c_str() is owned by the s object, so you can't access it beyond the lifetime of s (or after calling any non-const function on s).
It "works" and has no observable effect because you don't use t.
Nah, it's perfectly fine, you could print out t and it would work exactly as expected. But what latent said is very legit, once s goes out of scope, t is going to be undefined.
So, this code will cause an undefined behavior, but sadly won't explode, so it will be a bitch to track down.
#include <string>
using namespace std;
const char * GetString()
{
string s = "bob";
return s.c_str();
}
int main()
{
printf("%s",GetString());
return 0;
}
While this code would be perfectly fine, as s persists:
#include <string>
using namespace std;
const char * GetString(string & str)
{
return str.c_str();
}
int main()
{
string s = "bob";
printf("%s",GetString(s));
return 0;
}
[quote name='iMalc' timestamp='1319177105' post='4874931']
It "works" and has no observable effect because you don't use t.
Nah, it's perfectly fine, you could print out t and it would work exactly as expected.
[/quote]
I think what he meant is that it "works" as in it's perfectly well defined and valid C++ code, but it accomplishes absolutely nothing so it doesn't really "work," in the sense of accomplishing something.
But yeah, what Serapth said is a perfect example of the quirks of C++ that make it so easy to introduce a bug that is so bloody hard to track down. It's valid, but just be careful how you do it.
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