Learning C: Writing a Linked List Problem

[sneez]

Im trying to learn some c after java.

Im writing a linked list and i have this problem

from main i call : add( (x *)&head, (x *)&tail)

inside the function add i try to get input from user

blah balh
strcpy(tail->name, gets(str));


why does the program goes to next line without letting me to input the name ?

i hope I am clear if not let me know

sneez

 

[gerben]

probably there is still a newline in the buffer, to find out you can do this:

blah balh
printf("ABC %s DEF", gets());
strcpy(tail->name, gets(str));

if ABC and DEF get printed on different lines you know for sure

 

[TenaliRaman]

*agrees with gerben*
This has happened with me one times too many.
If you are sure that at no point in your program, a new line is making its way in, then try using fflush(stdin) . Ofcourse for some reasons, this is not guaranteed to work.

-- AI

 

[dduardo]

Just a note: Whenever doing strcpy operations make sure to check bounds. This is how you create buffer overflow problems.

 

[sneez]

thanx a lot

one more seemly unrelated question.

char *str="blabh";
Why tail->name=str //does not work ?

why do i have to use the strcpy ?

 

[gerben]

tail->name and str are both pointers (char *).
so
tail->name = str;
means assign the adress that str points to to the pointer variable tail->name, so that that pointer variable points to the same adress. So you will get a problem when the memory that str points to gets freed.

Probably both pointers (str and tail->name) already have some declared memory to which they point and you just want to copy the content of the memory to which str points to the memory that tail->name points to.
You could do this:
*(tail->name) = *str;
but that will only copy the first character of the string, so you should use strcpy()

 

[master_coda]

Just a note: Whenever doing strcpy operations make sure to check bounds. This is how you create buffer overflow problems.

Of course you should also never use gets for the same reason.

 

[tino lin]

thanx a lot

one more seemly unrelated question.

char *str="blabh";
Why tail->name=str //does not work ?

why do i have to use the strcpy ?

first of all str should be declare:
char str[] = "blabh"
to ensure the compiler generate the correct code to allocate memory for the array and not just a char pointer.

tail->name = str;
tail->name and str point to the same data in memory, if str is a local variable then that memory is gone when the function exit and tail->name is no longer valid. that is why you need you need to copy the data into tail->name location. Another danger of tail->name = str is that now you just lost the memory location of where tail->name was pointed to earlier, and the memory location allocate for tail->name is now loss. You just have cause a memory leak.