Dear all:
Following is the program I wrote:
But the output does not print the string I entered:Code:#include <stdio.h> int main () { char *string; printf("Enter something: "); scanf("%s", string); printf("The string you entered is %s\n", string); return 0; }
Enter something: string
The string you entered is (null)
If I change "char *string" to "char string[10]", it works fine, can anyone help me, thanks in advance
I realised that I haven't initialised the string pointer, if I do the following, it works.
Code:#include <stdio.h> int main () { char string[10]; char *strptr=string; printf("Enter something: "); scanf("%s", strptr); printf("The string you entered is %s\n", string); return 0; }Originally Posted by barramundi9
Last edited by barramundi9; 05-31-2011 at 03:45 AM.
That's because char*string does not allocate memory space for your string... it's merely a pointer to nothing.
C has no string type and is only marginally aware of text in programs. C-Strings are actually nothing but arrays of char sized integers. The string library functions (incl. scanf() and printf() ) merely treat them as strings by writing a 0 at the end of the used space in the char array.
If you want to use char *string ... you also have to call malloc() to set aside space for your char array and then free() it when you're done with it.
EDIT: I see from your second post you found an interesting solution... but in that case why not write to the char array directly? You can simply use scanf("%s", string); or use scanf("%9s",string) if you're (appropriately) worried about buffer overflows. An array's name is simply a pointer to it's first element, so you can work it directly with any of the library functions.
Last edited by CommonTater; 05-31-2011 at 03:51 AM.
Hmm ...
I just tried the same codes on my windows machine using Cygwin + Mintty and it worked!!
Really weird, any idea as to why it didn't work on my linux machine but it worked on my Cygwin on windows???
Thanks
What doesn't work? what happened?
Be specific.
You can only input 9 chars. if using #2 code
Sorry, I was referring #1 code
On my Linux server, the output was :Code:#include <stdio.h> int main () { char *string; printf("Enter something: "); scanf("%s", string); printf("The string you entered is %s\n", string); return 0; }
Enter something: string
The string you entered is (null)
But it worked (printed the string) on my windows pc with Cygwin+Mintty, maybe it has something to do with the compiler?
Undefined behavior. don/t play with it.
You aren't going to learn anything
No, it didn't work... Windows just didn't error off on it (and it should have).
Yes, it did because the string was printed, that was why I thought it was weird but I thought what Bayint Naung said was right, not worth checking out ...
Today's C lesson is: Correct output does not make a correct program.
Quzah.
Hope is the first step on the road to disappointment.
And when it turns out that uninitialized pointer is aimed directly at your program's stack, overwriting crucual data from other parts of the program? What then?
In a trivial example like yours it is likely you will get away with using uninitialized memory --that is, memory you don't own-- without any noticeable consequences... try that in a larger project, oh, lets say your company's inventory tracking, and watch what happens : "Hey Gerald, why does it say we have 430,006 of those chrome widgets?"
Er ... what I mean is that I'll give up on researching this technical glitch and forever initialise my pointers from now on ...
Actually your compiler will give warning. if you try to use un-initialize variables.
Don't you mean "compiler should give warning" ... whether it does or not depends on his compiler and the warning levels he's chosen...
You'd be surprised how many people simply turn off the warnings instead of fixing their code.
It's not a technical glitch... it's a difference between OSs ... and it's a part of the C language that you must somehow initialize all variables before using them... It won't do it for you.
