Thread: Questions..

Originally Posted by argv

by the way, is: "char *name" the same as "char name[]";

Nope! the former is a pointer while the latter is an array.

Originally Posted by argv

I dunno, but I like 'int* p' for some wierd reason. I guess as long as your consistent, it doesn't matter.

I think there's a link in the faq section on programming style which would be helpful.

Originally Posted by laserlight

In this context, no. The former declares name as a pointer to char, the latter declares name as an array of char with the size determined by the initialiser.

ok, so just so I'm clear, did you mean "less than" when you wrote this? (it seems like you would mean "more than"?

Originally Posted by laserlight

Somewhat, except that you do not know if argv[1] actually has less than 20 characters, so strcpy() risks buffer overflow.

Unless that is something with the strcpy function copying a string that is smaller than the array. But I thought if I make an array 20, then it will just copy whatever is 20 or less.

This is alot less automatic than those higher level languages.

Originally Posted by itCbitC

Nope! the former is a pointer while the latter is an array.

I think there's a link in the faq section on programming style which would be helpful.

Oh, thanks. I added that to my favorites list. And now I know what the hell a "foo" is I gues. cool.

Originally Posted by laserlight

strcpy() copies from the source to the destination until a null character is encountered. So, suppose the destination array is an array of 20 characters, including one character reserved for the null character. Supposed argv[1] contains all the letters of the English alphabet. There is not enough space to store the contents of argv[1] in the destination array, so strcpy() will keep copying, even if it means writing into memory that it should not be writing to.

Ok, gotcha. thank you.