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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
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.
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.
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.
Then how does this works?Quote:
Originally Posted by laserlight
Code:char *s="hello";
s="bye";
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*.Quote:
Originally Posted by BEN10
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.Quote:
Originally Posted by BEN10
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?Quote:
Originally Posted by hk_mp5kpdw
In the first example, it is at the first position in the memory allocated for the array.Quote:
Originally Posted by BEN10
In the first example, the first element (and indeed all the elements) of the array exists. Its value is another matter.Quote:
Originally Posted by BEN10
s is an array, not a pointer. When s is converted to a pointer, that pointer points to the first element of s.Quote:
Originally Posted by BEN10
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.
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 :D...
I dont know how your compiler ran the code but my error wasQuote:
Originally Posted by $l4xklynx
Quote:
The variable 's' is being used without being defined.
Do not use gets:
http://sourceforge.net/apps/mediawik...php?title=Gets