Halo all,
I'm posting here after a long time. During this time I was just sharpening my pointer skills but as pointers are the most "easiest" part of C I got stuck at this one.
Why does gets(although it's not advisable) not work with the second one but with the first.
Code 1:
Code 2:Code:char s[5]; gets(s);
As I see it s points to first element in both the cases, and both of them not initialized at all. So s doesn't points to a particular address initially.Code:char *s; gets(s);
Thanks
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
Declaring a pointer does not implicitly create something for that pointer to point at.
In "Code 2", s is an unitialised pointer, which means it points at some random memory location. gets(s) therefore overwrites that random area of memory. Formally, the result is undefined behaviour - anything is allowed to happen (a program crash is a common result, but other results are possible).
"Code 1" will exhibit undefined behaviour if the data read on a line from stdin exceeds 4 characters in length - because it will run past the end of the array. If the input on a line does not exceed 4 characters, all is fine.
Last edited by grumpy; 06-23-2009 at 07:13 AM.
Here's my real doubt. Isn't 's' an uninitialized pointer in code 1 coz 's' is the base address of the array. Also I changed my code 2 like this:
Code 2:
[CODE]
char *s="hello";
gets(s);
[\CODE]
But it still gives me run time error. What I think here is 's' is now initialized pointer so it should be working properly.
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
Well, now the problem is that s is a pointer to the first element of a string literal, thus you cannot modify what s points to without resulting in undefined behaviour.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Then how does this works?
Code:char *s="hello"; s="bye";
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
You are changing the pointer, not what the pointer points to. Oh, and if s is supposed to point to some character in a string literal, it should be a const char* rather than a char*.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
You said it yourself: "'s' is the base address of the array" Therefore, 's' isn't uninitialized, it represents the address of the first element of the array in this context.
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It sounds funny but where's the first element of the array? I mean 's' points to the first element of the array but the first element is not present so in effect 's' points to a random memory location just like char *s does. So how come gets work with it properly?
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
In the first example, it is at the first position in the memory allocated for the array.
In the first example, the first element (and indeed all the elements) of the array exists. Its value is another matter.
s is an array, not a pointer. When s is converted to a pointer, that pointer points to the first element of s.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
Thanks laserlight. You cleared a whole lot of doubts from my mind. Now I get exactly what's the differnce between char *s and char s[5]. Thanks once again.
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
I tried to use your code using codeblocks with mingw and it worked... I wonder what error you're encountering.?
edit: nevermind it's now resolved...
Last edited by $l4xklynx; 06-23-2009 at 09:15 AM. Reason: nvm
I dont know how your compiler ran the code but my error wasI tried to use your code using codeblocks with mingw and it worked... I wonder what error you're encountering.?
edit: nevermind it's now resolved
The variable 's' is being used without being defined.
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
Do not use gets:
http://sourceforge.net/apps/mediawik...php?title=Gets
