@JesseJames You're welcome. Yeah so I would check that "http://" thing on your router. If you can connect, then the connection between your router and your computer is fine and it is something else. But if you can't connect to that address then as siggy said it is probably a device driver thing. Use that wifi you can get from the cell phone to run the driver update wizard in your system settings.

@ThisIsChris I sure will thanks bro appreciate it!

@JesseJames yw, let us know how it goes!

So I got home tried the http thing and it worked and now it's working just fine.

@ThisIsChris

@JesseJames Cool, thanks for the update! Networking can be a finnicky thing. If that happens in the future I would try visiting that address again to see if that knocks it into shape

Is it possible that fixed something? If so what could it be?

There's a few things it could have been, all of them "bugs". When you connect to the internet, traffic goes from your computer to your router to your modem to your ISP, and then out to the rest of the world. Anywhere along that chain there could be problems getting traffic in or out. By connecting to that numerial address on your router, you were connecting to a website that is just on your router, so the traffic didn't have to go all the way to your ISP. In other words, you were *just* testing the connection between your computer and your router. This was simpler for the router then having to get traffic from the ISP and forward it to your computer, so by doing a simpler task it may have just reminded your router how to connect correctly to your computer, and then it was able to do the more complex task of forwarding you traffic from the real internet correctly

Ah. Just when I think I'm fairly good with computers. Lol.

haha to be fair without the manufacturer's code it's hard to know *exactly* what's going on, you just have some broad ideas of how things work and try poking around in them.

Does a hammer work for poking around?

lol

@everyone
```c
char c[] = "Str";
c[1] = 'r';
printf("%s\n", c); //Output: Srr
```
How to get this to work with an array of strings? I have an array of strings (char*) and I need to modify individual characters of each string. I'm getting segfaulted because I had type `char**`. Now I know `char[]` works for individual strings but how can I extend that to arrays of them? I'm writing this in C btw.

@JesseJames what do you think?

@John O - obviously I know the answer but I think the outcome would be satisfying.

Ball peen or claw?

Hmm that's a tough one.

You not happy with StackOverflow answer, @sigruna14 ?

1st answer here is easier than me writing out pointer thingy data structure: https://stackoverflow.com/questions/1088622/how-do-i-create-an-array-of-strings-in-c

@Why Tea I'm not sure how that addresses what I'm trying to do here.

Then maybe I don't understand the question. "How do I create an array of strings" is your question, and the second answer in that StackOverflow pretty much covers it.

For multiple cases.

No, I know how to create an array of strings. My question is how to modify individual characters of the strings within said array.

The answer to that is in there if you read carefully.

```c
char a[2][14];
strcpy(a[0], "blah");
strcpy(a[1], "hmm");```

You're talking about this? Wholesale string copy?

Are you wanting to hit the individual characters in the string and substitute? I think I did miss the gist of what the question is. You have variable length strings, in an array structure (pointers to a bunch of strings) but you want to change just individual characters at will?

It is an array of three strings, each of which is three characters long. It starts out with each string being `"000"`. I need to be able to reach in and change the zeroes to ones where necessary.

Aha.

But you want to reach in and "touch" each one individually, at will.

Pointer math

Yes.

Add data type size to array pointer to access each char

Assuming your language won't let you index the array like a normal person

That's all indexes are anyway. Array pointer + index * data type size

Well I mean, obviously it won't.

So do like `*array + (sizeof(char) * 2)`?

Yes

@Perihelion - CA Nope, "lvalue required as left operand of assignment"

I think you have to use &array to let it know you're treating the array like a bit literal

Ok do this for c++ *(array+2) to get value