array of pointers to array of strings

Question

we can declare array of strings the following way:

char *words[] = { "abc", "def", "bad", "hello", "captain", "def", "abc", "goodbye" };

Now I have an existing array of strings suppose:

   char strings[4][20];
   strcpy(strings[0], "foo");
   strcpy(strings[1], "def");
   strcpy(strings[2], "bad");
   strcpy(strings[3], "hello");

I want to do something like this:

   char *words[4];
   for ( j = 0; j < 4; j++)
   {
      words[j] = &strings[j];
   }

So that words would have the same structure as if I have defined it as in the beginning. Do you guys know how to do that?

What is the actual issue you are trying to solve? How the compiler lays it out "in the beginning" is probably a single chunk of memory. — John3136
– John3136, Commented Nov 29, 2013 at 4:09

perreal · Accepted Answer · 2013-11-29 04:08:56Z

3

No need for the address of:

char *words[4];
for ( j = 0; j < 4; j++)
{
   words[j] = strings[j];
}

answered Nov 29, 2013 at 4:08

perreal

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1 Comment

caf Over a year ago

It may be useful to note that another way of writing this is words[j] = &strings[i][0]; - the pointers are to the first character of the arrays that contain strings, not to the arrays themselves.

Havenard · Accepted Answer · 2013-11-29 04:17:06Z

0

char *words[] = { "abc", "def", "bad", "hello", "captain", "def", "abc", "goodbye" };

And

char *words[8];
words[0] = "abc";
words[1] = "def";
words[2] = "bad";
words[3] = "hello";
words[4] = "captain";
words[5] = "def";
words[6] = "abc";
words[7] = "goodbye";

Are the same things.

In both cases words is an array of pointers to chars. In C, strings are pointers to chars, literally. Same thing.

char *a[] = { "abc", "def", "bad", "hello", "captain", "def", "abc", "goodbye" };
char *b[8];
int i;
for (i = 0; i < 8; i++)
    b[i] = a[i];

Should work just fine too.

You only have to keep in mind that those are string literals, so they are read only. It's not some buffer you can modify.

And even if you use buffers, theres another point to observe:

char a[4][20];
char *b[4];
int i;

strcpy(a[0], "foo");
strcpy(a[1], "def");
strcpy(a[2], "bad");
strcpy(a[3], "hello");

for (i = 0; i < 4; i++)
    b[i] = a[i];

strcpy(b[0], "not foo");

printf("%s", a[0]);
// prints "not foo"

By modifying a string in b you modified the string in a aswell, because when you do b[i] = a[i] you are not copying the string, only the pointer, the memory address where its stored.

edited Nov 29, 2013 at 4:17

answered Nov 29, 2013 at 4:11

Havenard

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2 Comments

maximilliano Over a year ago

ye, but i have char a[8][20]; and char * b[8]. When I tried that it gives segmentation fault

Havenard Over a year ago

In this context you cannot copy from b to a because the pointers in a cannot be modified. a is a matrix.

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