char array vs. char pointer

Question

When receiving data through a socket using recv, I've noticed that, with:

char buffer[4];
memset(buffer, 0, 4);
recv(socket, buffer, 4, 0);

I receive

mesgx��

"mesg" being what I sent, with some random characters appended.

If I use

char * method = (char *) malloc(4);
memset(buffer, 0, 4);
recv(socket, buffer, 4, 0);

instead, I receive

mesg

So there's no random stuff appended to my string. I figured out that if I use char[5] instead it works as well, but I do not really understand why. Does malloc(4) really allocate 5 bytes, the fifth being a NUL?

Greg Hewgill · Accepted Answer · 2009-07-26 22:08:11Z

16

The call to malloc(4) really only allocates four bytes. It was just coincidence that the next byte in memory happened to be a NUL, which terminated your string for you.

In the case where you allocated char buffer[4] on the stack, the next byte happened to be an 'x' followed by some other stuff, so your string just continued until it found the next NUL byte.

Socket functions deal with bytes only, and don't treat the bytes as strings with a trailing NUL or anything extra. You get exactly what you ask for.

answered Jul 26, 2009 at 22:08

Greg Hewgill

1.0m192 gold badges1.2k silver badges1.3k bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Dirk is no longer here · Accepted Answer · 2009-07-26 22:02:45Z

6

You need 5 characters to be properly NULL terminated. The terminating NULL counts as one, so if we need N characters, allocate N+1. Or conversely, for an allocation of N you have N-1 available for your content.

answered Jul 26, 2009 at 22:02

Dirk is no longer here

370k60 gold badges668 silver badges742 bronze badges

Comments

Henk Holterman · Accepted Answer · 2009-07-26 22:10:52Z

3

I suspect the difference is a coincidence. When you use buffer as a string, for example in printf(), it will be read past it's limit until a '\0' is found.

You should use buffer[5] or malloc(5) in both cases. The memset() shouldn't be necessary, better put a buffer[4] = '\0' after the recv().

answered Jul 26, 2009 at 22:10

Henk Holterman

276k33 gold badges353 silver badges540 bronze badges

Comments

Lara Bailey · Accepted Answer · 2011-10-07 10:24:18Z

2

You can't possibly have received more than 4 chars as you only asked recv for a maximum of 4 bytes to be placed into your buffer. You should check the return value of recv to see how many bytes were actually returned.

I suspect that the issue is that you are not being careful to only output 4 chars from whatever routine is generating the output. One way of displaying the initial contents of a possibly non-null terminated char buffer is this.

printf("%.4s\n", buffer);

A complete recv call snippet might be:

#define MAX_BUF_LEN (512)
char buffer[MAX_BUF_LEN];
ssize_t count = recv(socket, buffer, MAX_BUF_LEN, 0);

if (count > 0)
    printf("%.*s\n", count, buffer);

edited Oct 7, 2011 at 10:24

answered Jul 26, 2009 at 22:09

Lara Bailey

801k110 gold badges650 silver badges667 bronze badges

Collectives™ on Stack Overflow

char array vs. char pointer

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related