At 5:13 PM -0800 12/13/02, Paul R. wrote:
No. "argvP[0]" is a "char*", so "&argvP[0]", is a "char**", which is what the updated function wants. Doing what you suggest should work, too; the two are equivalent.BTW, your "AddArg(&argvP[0], "This");" doesn't seem right, either. Passing argvP, argvP+1, etc. to AddArg (char **argvP ...) would work, I should think. Aren't you making a pointer pointer pointer?
However, I *should* point out that when I post sample snippets, I generally just type them into my mail program and compile them in my brain. I'm not like John Marshall, who will also test the snippet out in at least three different compilers first, and then run it by the ANSI committee. Or Ben Combee, who will try it out in CodeWarrior, find a codegen bug in the compiler, fix it, and then also write an article about it on <http://www.palmoswerks.com>.
-- Keith
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