find the bugs

[ak]

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

const char *getName(const char *c) {
std::string name = lookupName(c);
if (name == NULL)
return "Anonymous" ;
return name.c_str();
}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;
if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);
return 0;
}

MY SOLUTION : #bug 1: name==NULL is illegal

#bug 2: name=getName(c) should be included in
the 'if' statement so that NULL will
not be
passed to getName()

The above was my solution...what do you all experts think?
Am i right or have i missed something?

Cheers,
ak

[copx]

"ak" <aj**********@g mail.comschrieb im Newsbeitrag
news:11******** **************@ y80g2000hsf.goo glegroups.com.. .

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

const char *getName(const char *c) {
std::string name = lookupName(c);

One bug found: it's C++ :P

[snip]

[=?ISO-8859-1?Q?Bj=F8rn_Augestad?=]

ak wrote:

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

You're posting C++ code to comp.lang.c, not a great idea. Try
comp.lang.c++ instead.

>
const char *getName(const char *c) {
std::string name = lookupName(c);
if (name == NULL)
return "Anonymous" ;
return name.c_str();

<OT>
One bug: The function returns a pointer to mem which will be freed by
the std::string dtor.
</OT>

}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;
if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);

Bug 2: If argc < 2, then you call printf() with name==NULL.

return 0;
}

MY SOLUTION : #bug 1: name==NULL is illegal

If so, it is a syntax error and not a bug.

HTH
Bjørn

[snip]

[Richard Heathfield]

ak said:

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

const char *getName(const char *c) {
std::string name = lookupName(c);
if (name == NULL)
return "Anonymous" ;
return name.c_str();
}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;
if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);
return 0;
}

foo.c:1: warning: no previous prototype for `getName'
foo.c: In function `getName':
foo.c:2: parse error before `:'
foo.c:3: `name' undeclared (first use in this function)
foo.c:3: (Each undeclared identifier is reported only once
foo.c:3: for each function it appears in.)
foo.c:3: `NULL' undeclared (first use in this function)
foo.c:2: warning: label `std' defined but not used
foo.c:1: warning: unused parameter `c'
foo.c: In function `main':
foo.c:9: `NULL' undeclared (first use in this function)
foo.c:14: warning: implicit declaration of function `printf'
make: *** [foo.o] Error 1

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

[Martin Ambuhl]

ak wrote:

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

const char *getName(const char *c) {
std::string name = lookupName(c);

^^^^^^^^(1) ^^(2)
(1) syntax error
(and the label 'std:' is unused, but that's not an error)
(2) undefined identifier (which also makes c unused, but that's not an
error)

That's two. Do I need to find more? Ok.

if (name == NULL)

^^(3)
(3) undeclared identifier

return "Anonymous" ;
return name.c_str();
}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;

^^(4)
(4) undeclared identifier

if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);

^^^^(5)
(5) variadic function printf() used with out a declaration in scope

return 0;
}

MY SOLUTION : #bug 1: name==NULL is illegal

If 'name' and 'NULL' are defined, 'name == NULL' perfectly legal

#bug 2: name=getName(c) should be included in
the 'if' statement so that NULL will
not be
passed to getName()

Who cares? The example is hopeless broken without looking to the
'logic' of the code.

>
The above was my solution...what do you all experts think?
Am i right or have i missed something?

You are wrong. And you at least missed knowing the name of the language
you are using. The different language C++ has its own newsgroup.

[ak]

On May 4, 9:05 am, Bjørn Augestad <b...@metasyste ms.nowrote:

ak wrote:

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

You're posting C++ code to comp.lang.c, not a great idea. Try
comp.lang.c++ instead.

const char *getName(const char *c) {
std::string name = lookupName(c);
if (name == NULL)
return "Anonymous" ;
return name.c_str();

<OT>
One bug: The function returns a pointer to mem which will be freed by
the std::string dtor.
</OT>

}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;
if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);

Bug 2: If argc < 2, then you call printf() with name==NULL.

return 0;
}

MY SOLUTION : #bug 1: name==NULL is illegal

If so, it is a syntax error and not a bug.

HTH
Bjørn

[snip]

Sorrry for posting this question on c.l.c will remember that.
Thanks
AK

[Eric Sosman]

ak wrote On 05/04/07 11:58,:

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

const char *getName(const char *c) {
std::string name = lookupName(c);

#001: Syntax error.

if (name == NULL)

#002: No declaration for name.

#003: No declaration for NULL.

return "Anonymous" ;
return name.c_str();

#004: No declaration of name as a struct or union
type with a c_str element.

}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;

#005: No declaration for NULL.

#006: No declaration for NULL.

if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);

#007: No declaration for printf().

return 0;
}

MY SOLUTION : #bug 1: name==NULL is illegal

#bug 2: name=getName(c) should be included in
the 'if' statement so that NULL will
not be
passed to getName()

The above was my solution...what do you all experts think?
Am i right or have i missed something?

#010: Mistaking That Other Language for C.

--
Er*********@sun .com

[Beej Jorgensen]

Richard Heathfield <rj*@see.sig.in validwrote:

>foo.c:1: warning: no previous prototype for `getName'
foo.c: In function `getName':
foo.c:2: parse error before `:'

I'm not sure this is the answer the interviewers were looking for. :)

-Beej

[Joe Estock]

ak wrote:

Recently at an interview i was asked the following question :

Assuming the function lookupName is defined, what's wrong with this
code (hint: 2 bugs)?

I'm presuming the required headers were omitted for brevity (which,
generally in interviews they are since the focus is primarily on the
code syntax and/or logic).

>
const char *getName(const char *c) {
std::string name = lookupName(c);

(1) std::string is not a valid statement in C.

if (name == NULL)
return "Anonymous" ;
return name.c_str();

(2) Technically speaking you would generally receive the diagnostic
"Request for member not in a struct or union" or something similar since
"name" was not declared. Regardless of the diagnostic message this is,
in fact, the second error that the interviewer was looking for.

}

int main(int argc, char *argv[]) {
const char *name = NULL, *c = NULL;
if (argc >= 2)
c = argv[1];
name = getName(c);

printf("My name is %s\n", name);
return 0;
}

MY SOLUTION : #bug 1: name==NULL is illegal

How so? You're comparing a pointer to NULL which is perfectly legal.

>
#bug 2: name=getName(c) should be included in
the 'if' statement so that NULL will
not be
passed to getName()

Why? Both of the following statements would yield the same results,
however the first one is most certainly easier to maintain and read:

if (name == NULL)
{
/* code here */
}

if ((name = getName(c)) == NULL)
{
/* code here */
}

Note the additional set of parentheses around the function call. Passing
a NULL value to a function /shouldn't/ be an issue provided that the
function were written to handle NULL values (as it should).

>
The above was my solution...what do you all experts think?

I'm no expert, but I know my way around the language and I'll offer my
two cents. Whether you choose to listen or not is another story, however.

Am i right or have i missed something?

Cheers,
ak

No, you are not right and yes you have missed something. In fact you
have missed many "something" 's. I strongly suggest that you study the
language in more detail before attempting to apply for another position
in your chosen field. I realize that you did not write the above
fragment of code, however you also didn't analyze it correctly either.

Some interesting topics to focus on would be operator precedence,
general logic, as well as familiarizing yourself with the standard C
library and learning the keywords used in the language. Once you are
comfortable with these topics you could then focus on reading and
writing code. Pointers are also another area that you should focus on,
however I think that you should first get a grasp on the fundamentals of
the language (and programming in general, really).

[Christopher Benson-Manica]

Joe Estock <je*****@nospam nutextonline.co mwrote:

ak wrote:

std::string name = lookupName(c);

MY SOLUTION : #bug 1: name==NULL is illegal

How so? You're comparing a pointer to NULL which is perfectly legal.

name is not a pointer. Such are the perils of posting C++ questions
to comp.lang.c. OP is correct.

Some interesting topics to focus on would be operator precedence,
general logic, as well as familiarizing yourself with the standard C
library and learning the keywords used in the language.

Given that the code (and the errors) are specific to C++, this advice,
while obviously sound, is unlikely to help OP pass a C++ exam.

--
C. Benson Manica | I *should* know what I'm talking about - if I
cbmanica(at)gma il.com | don't, I need to know. Flames welcome.