[c] reading a string on second letter

[saiz66]

Hi. I wanted to know how I could start off a string from the second letter. An example would be better:

eg.

char mybuf[100];

mybuf = "Hi.";

I want it so it is:

mybuf = "i."

So basically I am starting off at the 2nd spot. Thanks!

Also i want to add that i know how to do it when it is

char * mybuf;

but is it possible to do it with

char mybuf[100]?

[kjordan2001]

You should be able to use

char * strncpy ( char * dest, const char * src, sizet_t num )

Do something like:

strncpy(dest,&mybuf[1],strlen(mybuf)-2);

or

strncpy(dest,mybuf+1,strlen(mybuf)-2);

may work as well.

[Mouton]

Do it the exact same way you would with a char*: use "myBuf+1".

A char[] is the exact same thing a char*, except that with a char* u need to allocate and de-allocate memory yourself; both are pointers...

btw kjordan2001: no need to use a strncpy... a strcpy would work just fine since it would copy until the eos char (\0)...

strcpy(dest,myBuf+1);

strcpy(dest,&mybuf[1]);

[ilmcuts]

  Mouton said:

no need to use a strncpy... a strcpy would work just fine since it would copy until the eos char (\0)...

strcpy(dest,myBuf+1);

strcpy(dest,&mybuf[1]);

584723600[/snapback]

But that might be a buffer overflow. You don't know if the string fits into the buffer. If it is filled to the last byte, your strcpy would overflow the buffer by one. If you think that doesn't matter - remember how every other critical security hole involves unchecked buffers?

At the same time...

  kjordan2001 said:

strncpy(dest,&mybuf[1],strlen(mybuf)-2);

or

strncpy(dest,mybuf+1,strlen(mybuf)-2);

584723180[/snapback]

Those are subtly broken as well, with at least two bugs. They are probably just oversights: Using strlen(myBuf)-2 here doesn't make sense. Consider what happens if there are 99 non-null characters plus the '\0' at the end in myBuf. strlen() returns 99. You pass 97 as the buffer size to strncpy. It will thus truncate the string.

And it will reveal the second bug: strncpy is not guaranteed to zero-terminate the buffer. If there is not enough room for the original string in the destination buffer, the last character in the destination buffer will be the last from the source buffer that could be copied, rather than '\0'. Thus people usually do the following as a workaround:

strncpy(target, source, sizeof(target));
target[sizeof(target)-1] = '\0';

You can imagine how easy it is to forget that of course... Thus some people instead ensure that the buffer is filled with zeros to begin with....

char target[100] = { 0 };

...and then use

strncpy(target, source, sizeof(target)-1);

So strncpy never touches the very last (zero) byte. That's most likely what you had in mind.

You probably meant to use sizeof() instead of strlen() as well. Hence my gut feel that it's just an oversight.

I'm not sure if strncpy's behavior is defined if source and target overlap anyway. Though I've seen and used this so many times that it apparently works at least with popular compilers/library implementations.

[Mouton]

  ilmcuts said:

But that might be a buffer overflow. You don't know if the string fits into the buffer. If it is filled to the last byte, your strcpy would overflow the buffer by one. If you think that doesn't matter - remember how every other critical security hole involves unchecked buffers?

I know it won't overflow because I defined the char arrays myself with the appropriate sizes.

No way u can buffer overflow if dest and myBuf are of the same size. And I wouldn't see why you would not make them the same size. Especially since he's using char arrays, and not pointers + mem allocation...

I think choosing the size of your arrays correctly is much easier to deal with than using strncpy then having to think/check for the \0...

[ilmcuts]

Scratch my entire post. It's what happens when you post at 6 AM in the morning. I just read over it again, it's rubbish. I apparently read str(n)cpy(buf+1, buf). Apologies for the trouble caused. :\

[kjordan2001]

  Quote

So strncpy never touches the very last (zero) byte. That's most likely what you had in mind.

You probably meant to use sizeof() instead of strlen() as well. Hence my gut feel that it's just an oversight.

Yeah, I meant sizeof. However, I was right on the -2 part, since he's excluding the first letter.

  Quote

btw kjordan2001: no need to use a strncpy... a strcpy would work just fine since it would copy until the eos char (\0)...

Yeah, that would work too, guess I always go the long hard way :wacko: