why strncpy not copy 4 character in c2 ?

JonB

@Qt-embedded-developer
And where does c2 point to in your code?

As a separate matter, depending on whether you have any further code, you may not see anything anyway. How about cout<<c2<<endl?

Qt embedded developer

@Christian-Ehrlicher i have edited the question. can you just let me know why what i expect not come in output.

i have seen that when i use char array it show output what i expect. but why its not possible with char* ?

JonB

@Qt-embedded-developer
Goodness, how does setting to c2 to NULL help? Do you not understand strncpy(), please read its documentation.....

i have seen that when i use char array it show output what i expect. but why its not possible with char* ?

Sorry, but you really need to read up on C basics....

Further, you will see it does not put a terminating \0 into the destination area. If cout<<s2 prints something it would be hell followed by an unknown stream of random characters.....

Qt embedded developer

@JonB

why this show correct output then ?

char* c= "hello world",c2[10];
memset(c2,0x00,sizeof(c2));
strncpy(c2,c,4);

cout<<c2;

JonB

@Qt-embedded-developer
Precisely because that avoids all the faults previously mentioned in your code!

char c2[10] allocates 10 bytes of storage which c2 "points to".

char *c2 allocates no storage for c2 to point to. You would either need to malloc() space, or set c2 to point to, say, a char c3[10] which does allocate storage.

You seem to think char c2[10] and char *c2 are "the same thing", which they are not.

Qt embedded developer

@JonB can you give example to put \0 in above example code ?

JonB

@Qt-embedded-developer
I really, really should not have to do this....

char *c = "hello world";
char c2[10];
strncpy(c2, c, 4);
c2[4] = '\0';
cout << c2 << endl;

One further thing: these days, using C++ (and also Qt), you really should no longer need to use C library functions like strncpy() (or malloc()). There are better, safer ways to do that, e.g. using C++ std::string or Qt QString (and new for malloc()). Same with e.g. QByteArray instead of memcpy/set().

Qt embedded developer

@JonB said in why strncpy not copy 4 character in c2 ?:

char *c = "hello world";
char c2[10];
strncpy(c2, c, 4);
c2[4] = '\0';
cout << c2 << endl;

sorry for not clearly say about code. i said about below code.

can you just let me know what change in below code i need to do to add '\0'.

char* c= "hello world",*c2;
c2= NULL;
memset(c2,0x00,sizeof(c2));
strncpy(c2,c,4);

JonB

@Qt-embedded-developer
No, I'm sorry, I have reached my limit as to what I consider suitable to write/explain for you on this Qt site. For your own good you need to (a) read up about basic C, (b) understand code and (c) write/adapt your own for really simple stuff instead of asking someone else to do it for you.

You have ignored what I said about c2 = NULL, and any understanding of allocating space.
You have not made any attempt to look at what I did write for you and plainly/easily apply it to your new case.
You have written (copied?) memset(c2,0x00,sizeof(c2)); and not understood how that would affect any need to "add '\0'.".

jsulm

@Qt-embedded-developer said in why strncpy not copy 4 character in c2 ?:

can you just let me know what change in below code i need to do to add '\0'.

YOU NEED TO ALLOCATE MEMORY FOR c2...
You was already told so here: "You would either need to malloc() space, or set c2 to point to, say, a char c3[10]".
Please try at least to understand what others are writing.

As others already pointed out: learn C/C++ basics.

Qt embedded developer

@jsulm @JonB Thank you

i got the success:

char* c= "hello world",*c2;
int n=5;
c2= NULL;

c2 = (char*) malloc(n*sizeof(char));
strncpy(c2,c,4);

cout<<c2; // this print the hell

JonB

@Qt-embedded-developer
Yeah, but why in the world you need such code with strncpy() & malloc() is quite beyond me....

P.S.
Technically your code is still not right. You will only copy hell, 4 characters, into the malloc()ed area, and that does initialize its memory allocation. The 5th character onward could be anything. You are "lucky" the cout<<c2 does not print helljgas978623i5tagsd534... :)

strncpy(c2,c,4);  // copies "hell", 4 characters, nothing more, does not add any terminator
c2[4] = '\0';  // need terminating '\0' here
cout<<c2; // this print the hell --- FOR SURE!

SimonSchroeder

@JonB said in why strncpy not copy 4 character in c2 ?:

Technically your code is still not right.

Also: There is a memory leak because there is no matching free() for the malloc().

Just because it works now, does not mean it is valid code. You need to make sure for every C-style string that there is a terminating '\0'. There is no other way to know the length of the string. Every C algorithm expects the terminating '\0'. The reason why this might work right now could be that you are compiling a debug version and this will initialize the memory with zeros. Once you compile a release version your application might suddenly crash. At least it is not guaranteed to work. Another reason that it works might be because you are lucky that there is a 0 in memory. Just one little change in code might get you a different memory allocation/access pattern and the 0 is gone.

Since you are already using cout this means you are using C++. C++ is here to help you, not to punish you. Use std::string instead of char*. Try to avoid new and malloc because you always might forget the corresponding delete or free. std::string and std::vector (and other containers) will gladly manage memory for you. If you really need a pointer, use a smart pointer (together with make_unique or make_shared, so you don't have to use new which might accidentally leave a memory leak).