C++ reference and NULL

[m.labanowicz]

Hi,

There is interesting example about NULL and reference usage in C++

typedef struct {
int a [2];
} s_t;
int bar(int v, s_t &s) {
return v ? s.a[0] : 0;
}
int main(int argc, char * argv []) {
s_t local;
s_t * ptr = &local;
ptr = 0;
((void)(argv));
return bar(argc - 1, *ptr);
}

Which line do you expect that APP crashes ?

Regards

 

[Barry Schwarz]

Hi,

There is interesting example about NULL and reference usage in C++

typedef struct {
int a [2];
} s_t;
int bar(int v, s_t &s) {
return v ? s.a[0] : 0;
}
int main(int argc, char * argv []) {
s_t local;

The contents of local are indeterminate. Even if you passed local to
bar() by reference, which you don't, the evaluation of local.a[0]
would cause undefined behavior.

s_t * ptr = &local;

Given the next line, what purpose does the initialization serve? Or
why not initialize it to 0 immediately?

ptr = 0;
((void)(argv));

What purpose does this line serve? argv is a pointer guaranteed to
have a valid value. Evaluating that address and then discarding the
result is effectively a no-op.

return bar(argc - 1, *ptr);

Any attempt to dereference a NULL pointer invokes undefined behavior.

}

Which line do you expect that APP crashes ?

Undefined behavior is not guaranteed to crash the app.

 

[Jorgen Grahn]

(e-mail address removed) wrote: ....

This is the line that puzzles me.

You see it in C code sometimes ... it makes at least gcc shut up
about possibly unused parameters.

I cannot explain the extra pairs of () though.

/Jorgen

 

[Luca Risolia]

typedef struct {
int a [2];
} s_t;
int bar(int v, s_t &s) {
return v ? s.a[0] : 0;
}
int main(int argc, char * argv []) {
s_t local;
s_t * ptr = &local;
ptr = 0;
((void)(argv));
return bar(argc - 1, *ptr);
}

Which line do you expect that APP crashes ?

$ cppcheck main.cpp
Checking main.cpp...
[main.cpp:12]: (error) Null pointer dereference

which is UB: the program will crash if you are lucky.