why is strncpy segfaulting here?

Calef13 · 12-28-2008, 10:49 PM

It's ok my question is already answered here: fscanf causes a SEGMENTATION FAULT in post #5. Sorry for the repost

Hi,

I am writing some code to parse a port range given as a string eg: "20-25". After I figure out where the first port ends and I try to copy it however I get a segfault:

Code:

if( ( dash = strchr(ports, '-')) != NULL )
    {
        /* port range given */
        int index = dash - ports;
        if(index < 0)
            index = -index;
        char *temp;
        fnshPort = atoi( (dash + 1) );
        strncpy( temp, ports, index ); /* here is the segfault */
        strtPort = atoi(temp);

Code:

Breakpoint 1, main (argc=3, argv=0xbfbf5e84) at starshine.c:59
59	        strncpy( temp, ports, index ); /* here is the segfault */
(gdb) print temp
$1 = 0xb80e9e00 "U\211�WVS�p\207"
(gdb) print ports
$2 = 0xbfbf7c13 "20-25"
(gdb) step

Program received signal SIGSEGV, Segmentation fault.
0xb7fee41c in strncpy () from /lib/libc.so.6

The pointer arithmetic is copied from a website, but it makes sense to me. What has me stumped is that if I declare temp as an array it works fine, both lines below fix the problem:

Code:

char temp[15];

Code:

char temp[index];

I tried the later one but gdb shows that it doesn't get declared any smaller than the first, which was my intention, is using a number the only decent way to do it? index is 2 for the above example of "20-25".

So my question is: why doesn't a plain old c string work?

tabstop · 12-28-2008, 10:57 PM

Because char temp[15] is a plain old C-string, and char *temp isn't?

More to the point, temp will point to a C-string, but you have to have one around to point to; right now you are pointing at New Jersey (some people claim that it's nowhere) and that's no place to try to write a string to.

temp[index] is valid in C99 -- but temp[2] is not nearly big enough to hold "20", which requires three characters.

gooey_kablooie · 12-28-2008, 11:00 PM

your temp pointer is pointing to some memory not allocated. if u read the man page for strncpy, it says that your destination should be a buffer i.e. u cant pass a pointer that has no memory allocated.

obviously if u declare an array it will solve the problem as it is now pointing to a legal block of memory.

cas · 12-29-2008, 03:27 PM

And do be careful with strncpy(). It doesn't necessarily create a string, despite its name; and in your code, it definitely won't create one (without some help).

The problem is that if strncpy() doesn't see a null character in the source string, it won't put one in the destination string (which means it's not actually a string at all). As far as I can see, your code will essentially be doing strncpy(temp, "20-25", 2). Since neither '2' nor '0' is a null character, no string will be created. An easy way to fix that is to do temp[index] = 0 after your strncpy().

12-28-2008, 10:49 PM	#1
Calef13 Registered User Join Date: Dec 2005 Location: german border Posts: 72	why is strncpy segfaulting here? It's ok my question is already answered here: fscanf causes a SEGMENTATION FAULT in post #5. Sorry for the repost Hi, I am writing some code to parse a port range given as a string eg: "20-25". After I figure out where the first port ends and I try to copy it however I get a segfault: Code: if( ( dash = strchr(ports, '-')) != NULL ) { /* port range given / int index = dash - ports; if(index < 0) index = -index; char temp; fnshPort = atoi( (dash + 1) ); strncpy( temp, ports, index ); /* here is the segfault / strtPort = atoi(temp); Code: Breakpoint 1, main (argc=3, argv=0xbfbf5e84) at starshine.c:59 59 strncpy( temp, ports, index ); / here is the segfault / (gdb) print temp $1 = 0xb80e9e00 "U\211�WVS�p\207" (gdb) print ports $2 = 0xbfbf7c13 "20-25" (gdb) step Program received signal SIGSEGV, Segmentation fault. 0xb7fee41c in strncpy () from /lib/libc.so.6 The pointer arithmetic is copied from a website, but it makes sense to me. What has me stumped is that if I declare temp as an array it works fine, both lines below fix the problem: Code: char temp[15]; Code: char temp[index]; I tried the later one but gdb shows that it doesn't get declared any smaller than the first, which was my intention, is using a number the only decent way to do it? index is 2 for the above example of "20-25". So my question is: why doesn't a plain old c string work? Last edited by Calef13; 12-28-2008 at 10:56 PM.*
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12-28-2008, 10:57 PM	#2
tabstop and the Hat of Guessing Join Date: Nov 2007 Posts: 8,740	Because char temp[15] is a plain old C-string, and char *temp isn't? More to the point, temp will point to a C-string, but you have to have one around to point to; right now you are pointing at New Jersey (some people claim that it's nowhere) and that's no place to try to write a string to. temp[index] is valid in C99 -- but temp[2] is not nearly big enough to hold "20", which requires three characters.
tabstop is offline

12-28-2008, 11:00 PM	#3
gooey_kablooie Registered User Join Date: Dec 2008 Location: Loch Ness Posts: 6	your temp pointer is pointing to some memory not allocated. if u read the man page for strncpy, it says that your destination should be a buffer i.e. u cant pass a pointer that has no memory allocated. obviously if u declare an array it will solve the problem as it is now pointing to a legal block of memory.
gooey_kablooie is offline

12-29-2008, 03:27 PM	#4
cas Registered User Join Date: Sep 2007 Posts: 372	And do be careful with strncpy(). It doesn't necessarily create a string, despite its name; and in your code, it definitely won't create one (without some help). The problem is that if strncpy() doesn't see a null character in the source string, it won't put one in the destination string (which means it's not actually a string at all). As far as I can see, your code will essentially be doing strncpy(temp, "20-25", 2). Since neither '2' nor '0' is a null character, no string will be created. An easy way to fix that is to do temp[index] = 0 after your strncpy().
cas is offline

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