strings of ASCII

As a newbie I was trying to declare an ASCII string ending with a "0", but got a compiler error.

Code:

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constant char STRING[7] = {"string",0};

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What is wrong?

Because when you initialize a string, you need to do it one character at a time.

Like this:

Code:

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const char STRING[] = {'s', 't', 'r', 'i', 'n', 'g', '0'};

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or you can do this...

Code:

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//Also, include <string.h> 
  char STRING[10];
  strcpy(STRING,"string");
  strcat(STRING,"0");

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Quote:

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Originally Posted by SlyMaelstrom

Because when you initialize a string, you need to do it one character at a time.

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Code:

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const char astringy[] = "You do?";

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Quzah.

Ok you don't... but you could.

Quote:

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Originally Posted by Andy_P

As a newbie I was trying to declare an ASCII string ending with a "0", but got a compiler error.

Code:

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constant char STRING[7] = {"string",0};

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What is wrong?

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you don't have to explicitly put the terminating 0 there, the compiler will do that for you. You also don't need the braces, nor specify the string size unless you want the character array to be larger than the initialization string.

Code:

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constant char STRING[] = "string";

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Quote:

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Originally Posted by Ancient Dragon

Code:

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constant char STRING[] = "string";

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And it's const, not constant.