Ben Bacarisse wrote:
> Hal Vaughan <hal@[EMAIL PROTECTED]
> writes:
>
>> Ben Bacarisse wrote:
>>
>>> Hal Vaughan <hal@[EMAIL PROTECTED]
> writes:
>>>
>>>> Hal Vaughan wrote:
>>>>
>>>>> Hal Vaughan wrote:
>>>> ...
>>>>> There's one problem that didn't show up until I fixed this one.
>>>>>
>>>>> I have my own library with this function:
>>>>>
>>>>> void parseargs(int, char*);
>>>
>>> This declaration does not match the definition you give later.
>>
>> Okay. I'm a bit confused. I have:
>>
>> int main(int argc, char* argv[])
>>
>> which leads me to believe argv is of type char*,
>
> Ah, no. In C and C++, the [] in a parameter list act almost exactly
> like a *. argv is and array of char pointers and thus becomes a
> pointer to a the first char pointer when it is passed to main. The []
> form is transitional but a little confusing.
>
>> so when I call
>>
>> parseargs(argc, argv)
>>
>> I would expect parseargs to have the same signature as far as the
params
>> go:
>>
>> map<string,string> parseargs(int count, char* args[])
>
> That is correct.
>
>> From what you say, it should be:
>>
>> map<string,string> parseargs(int count, char** args[])
>
> No. Your problem is in the declaration. You showed a line:
>
> void parseargs(int argc, char *argv);
>
> which is quite different.
So now that I have
map<string,string> parseargs(int count, char** args[])
how do I access the individual char[] arrays in args (which is argv being
passed to the function)? I can't do:
string arg;
arg = args[x]
I've tried that and it doesn't work. Do I need to dereference it or
something?
Hal
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