copy long strings in C

[Mark McIntyre]

My point was simply that I found your use of the term "crash"
confusing. As I understand the term, if the program quietly continues
running incorrectly, it hasn't "crashed"; a "crash" is a very visible
failure. (At least one other person made the same point.)

This usage of the term is consistent with the older real-world
meaning. If I'm driving down the highway and my brakes fail, the car
is not operating correctly -- but it hasn't "crashed" until it
actually runs into something.

I've several times encountered systems which were still running, but
no longer responsive to user input, didn't do what they were supposed
to, or otherwise behaved unsuitably. As far as I'm concerned, the app
had crashed, the fact that it didn't pop up a bluescreen, produce a
coredump or advise me via cutesy dialogs didn't make it any the less
crashed.

But whatever. Its amazing that a throwaway remark can generate such
sound and fury, and still signify nothing. Here's what I originally
said, and which has stirred up such ire:

char st1[1000],st2[650]; ..
st2[650] = '\0';

 

[Keith Thompson]

But whatever. Its amazing that a throwaway remark can generate such
sound and fury, and still signify nothing. Here's what I originally
said, and which has stirred up such ire:

char st1[1000],st2[650]; .
st2[650] = '\0';

st2 doesn't have a 651st element, so this will cause a crash of some
sort.

And my point is that it's just as likely to quietly do just what it
would have done if st2 actually had 651 elements. It's undefined
behavior; it may or may not cause any misbehavior at all, visible or
otherwise.

If you write past the end of an array, you *might* step on memory that
your process doesn't own, but it's just as likely to step on memory
that's part of another variable (with arbitrarily bad consequences) or
on a gap between variables (with, most likely, no consequences at
all).

Try this program:

#include <stdio.h>
int main(void)
{
char st1[1000],st2[650];
st2[650] = '\0';
printf("st2[650] = %d\n", st2[650]);
return 0;
}

For me, it simply printed "st2[650] = 0" followed by a newline. Does
it "crash" when you run it?

If you invoke undefined behavior, you cannot depend on the
implementation to catch it for you.

 

[Mark McIntyre]

And my point is that it's just as likely to quietly do just what it
would have done if st2 actually had 651 elements.

Unfortunately however, your point came across (to me) as:
"Ha, I'm feeling excessively pedantic and will make a big thing of the
fact that it needn't actually terminate your app" but perhaps I too
was feeling overly sensitive.

For me, it simply printed "st2[650] = 0" followed by a newline.

How do you know that this is all it did? Perhaps that byte was used by
a daemon which is now quietly writing logs to "\0ogfile.log". Or
perhaps your password is now "passw\0rd"....

Does it "crash" when you run it?

FWIW, where did I say that the app itself would crash? Of course, this
is just me being pedantic too.

If you invoke undefined behavior, you cannot depend on the
implementation to catch it for you.

Absolutely.

It seems to me however that your statement about what it simply did
for you, is kinda assuming /your/ OS has done just that !

End of pedanticism and anality I hope...

 

[Keith Thompson]

Unfortunately however, your point came across (to me) as:
"Ha, I'm feeling excessively pedantic and will make a big thing of the
fact that it needn't actually terminate your app" but perhaps I too
was feeling overly sensitive.

I genuinely did not understand what you meant, and I don't believe I'm
the only one. The plain meaning of what you originally wrote, as I
understood it, was that the program would certainly fail in some
visible way. I understand now that that's not what you meant.

For me, it simply printed "st2[650] = 0" followed by a newline.

How do you know that this is all it did? Perhaps that byte was used by
a daemon which is now quietly writing logs to "\0ogfile.log". Or
perhaps your password is now "passw\0rd"....

Because I know enough about the system I'm using to be reasonably
confident that things like that are extremly unlikely consequences of
that particular instance of undefined behavior. This knowledge is
off-topic, of course; the C standard makes no such guarantees.

FWIW, where did I say that the app itself would crash? Of course, this
is just me being pedantic too.

Ok, you said "this will cause a crash of some sort". Did it do so?

Absolutely.

It seems to me however that your statement about what it simply did
for you, is kinda assuming /your/ OS has done just that !

I don't follow your reasoning. Neither the OS, the C implementation,
nor the program itself "caught" the undefined behavior. The program
(I'm reasonably sure) simpy wrote a value to an unused byte. Are you
using "catch" in some sense that I'm not familiar with?

End of pedanticism and anality I hope...

I hope not. }

 

[Mark McIntyre]

I genuinely did not understand what you meant, and I don't believe I'm
the only one.

For which my apologies.

The plain meaning of what you originally wrote, as I
understood it, was that the program would certainly fail in some
visible way. I understand now that that's not what you meant.

Actually what I meant was "hey, don't do that, if you persist to do
it I guarantee at some time in your career, you'll become highly
embarassed, unemployed or possibly even dead, depending on the nature
of the crash and the circumstances surrounding it"

Ok, you said "this will cause a crash of some sort". Did it do so?

I've no idea as I'm not sitting in front of your computer. Have you
checked your phone bill yet? Is the digital TV playing up? Are you
being followed by the FBI?

I hope not. }

Well, at least in this context !