How to define an array of pointers to array of pointers in C?

Question

Lets say I have the following pointer arrays:

const char* m1[5] = { "bla", "asdada", "sadasde", "wrskm", "adjsad" };
const char* m2[5] = { "xxx", "yyy", "zzz", "uuu", "vvv" };
const char* m3[5] = { "lkkl", "kkk", "lkkl", "skl", "jkljkl" };
const char *m4[5] = { "one", "two", "three", "four", "five" };

I just want to represent the data above in a single array where each element represents one of the array above. like array of array of pointers to cons char structure(corrected me please if this saying wrong) I try to do the following but it doesn't work

char *(const char* exm[5])[4] = { &m1, &m2, &m3, &m4};

Help?

Some programmer dude · Accepted Answer · 2019-10-09 19:48:19Z

2

Remember that arrays naturally decays to pointers to their first element. For an array of pointers, that becomes a pointer to a pointer, i.e. const char ** for your arrays m1 to m4.

So you need an array of pointer to pointers:

const char **exm[] = { m1, m2, m3, m4 };

answered Oct 9, 2019 at 19:48

Some programmer dude

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