C - Array of pointers to string

Question

How to do in C - array of pointers to string? Because the string is represented as array of chars i tried to do this:(assumes every string is maximum 10 chars and the array size is 100)

char[10]* array[100];

but it is wrong

Any suggestion?

Take a look to your main function. There you have exactly that for the argv parameter (char** or char* argv[]). — Adriano Repetti
– Adriano Repetti, Commented Nov 5, 2013 at 16:58
One (often difficult) thing you need to wrap your head around: you can have array of pointers (here pointers to arrays of char, possibly of different length, or pointer can be NULL too). Then you can have array of arrays (here of char) that is two dimensional array, where each inner array exists (can't be NULL) and has fixed length ('\0'-terminated string can be shorter, with unused space after the 0). With strings, you usually want the first, array of pointers to arrays of char. — hyde
– hyde, Commented Nov 5, 2013 at 17:07

abelenky · Accepted Answer · 2013-11-05 17:00:08Z

2

As Adriano said, in C, the main function contains an array of strings:

int main(int argc, char* argv[])
{ [...] }

argv is an array of strings, and is properly declared.
Copy that. (and if you run into problems, ask a specific, detailed question)

answered Nov 5, 2013 at 17:00

abelenky

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olegarch · Accepted Answer · 2013-11-05 17:00:14Z

0

For static allocation:

char buf[10][100];

For dynamic allocation:

char *buf[10];

And assign thereafter:

buf[5] = strdup("Hello");

answered Nov 5, 2013 at 17:00

olegarch

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2 Comments

manuell Over a year ago

you mean char *buf[100]; don't you?

hyde Over a year ago

The "for dynamic allocation" version can be static too: const char *arr[] = { "str1", "str2" };

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