array of strings

OK listen, the -> syntax is just a shortcut for "(*myvar)."
And if you know what * is, then you know it dereferences a pointer. So in short, it's the same as using "." for accessing members, but for pointers instead.

Then not that this is wrong and bad:

Code:

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char* test = "My name is John and I love to surf"

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The proper code should be:

Code:

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const char* test = "My name is John and I love to surf"

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Because string literals are constant.

And I'm pretty sure you're using Visual Studio, so I'd also suggest you take a look at safer functions such as strcpy_s and the like.

You still seem to be confused by the -> operator. It doesn't have anything to do with what is on the right side but what is on the left side. temp->rule->target is likely to cause serious problems because temp is not a pointer it is a structure. So we have to start out with temp.xxxx

Now what member of temp are you looking for? temp.rule is a pointer to a rule_t It has not been allocated yet so you have some other problems. If you did have a rule_t allocated in temp by the name of "rule" then you could access it's members by temp.rule->target which is also a pointer that points off into never never land.

I really think you should back off and work on some simpler exercises until you REALLY understand memory allocation and pointers. Pointers to structures that contain pointers to other structures that have pointers to pointers are going to really mess you up unless you have a VERY good grasp on the subject matter.

well what if I have a pointer to that structure then I could do temp->rule->target right?

Yes, assuming that you have it all properly allocated.

Quote:

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Originally Posted by Elysia

Then note that this is wrong and bad:

Code:

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char* test = "My name is John and I love to surf"

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To clarify that, since I can't edit. It was a typo, and should say "note", not "not."
Typo highlighted in quote.

Quote:

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Originally Posted by -EquinoX-

well what if I have a pointer to that structure then I could do temp->rule->target right?

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Let's see it this way. There is no "->". There is only ".".
To get the contents of a pointer, you need to dereference it first. So *ptr gives you its contents, and not the address stored in the variable.
Then if it's a struct and has members, we use the "." to access those members.
However, *ptr.something is wrong because the "." operator has precedence over the "*" operator, which means you'll be trying to access members from the pointer itself and then dereference that members, which is wrong. Pointers don't have members.
The solution to that is to use parenthesises to prioritize the dereference operator like so (*ptr).something. This is awkward syntax and thus the "->" operator was born to do exactly that, so (*ptr).something becomes ptr->something.

Don't forget.