How to select an array of strings in C?

Question

I want to define different arrays of strings in C, which then can be e.g. selected depending on some other value, i.e. like following:

char foo[][10] = {"Snakes", "on", "a", "Plane"};
char bar[][10] = {"Fishes", "in", "a", "Lake"};
char *choice;
if (flag == 1) {
  choice = &foo;
} else if (flag == 2) {
  choice = &bar;
} 
printf("%s%s\n", choice[0] , choice[1]);

Expected result in case flag is 1:

Snakeson

Expected result in case flag is 2:

Fishesin

But the above code gives a segmentation fault error, while I tried differnt definitions for char, i.e. char* and char**. How to do it right? Is there a good tutorial on this matter, i.e. on pointers, arrays, what foo exactly is in the above example...

see man page of printf. You are using it in a very wrong way!try printf("%s%s\n", choice[0] , choice[1]); — Varvarigos Emmanouil, Jun 24 '13 at 11:08
@Maneolin - you are right - I updated the code. But error is still the same (`segmentation fault`). — Alex, Jun 24 '13 at 11:11
Thread http://stackoverflow.com/questions/2003745/pointer-address-in-a-c-multidimensional-array should really help you. — João Fernandes, Jun 24 '13 at 11:41

score 5 · Accepted Answer · answered Jun 24 '13 at 11:12

5

It's easier if you just use arrays of pointers:

int main(void)
{
  const char *foo[] = { "Snakes", "on", "a", "Plane" };
  const char *bar[] = { "Fishes", "in", "a", "Lake" };
  const int flag = 17;
  const char **choice = (flag == 1) ? foo : bar;

  printf("%s %s\n", choice[0], choice[1]);

  return 0;
}

The above prints

Fishes in

answered Jun 24 '13 at 11:12

unwind

391,730
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469
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Your suggestion seems to work! I will try it in the actual code. – Alex Jun 24 '13 at 11:18

Joni · Answer 2 · 2013-06-24T11:29:11.040

With arrays of char you need this:

char foo[][10] = {"Snakes", "on", "a", "Plane"};
char bar[][10] = {"Fishes", "in", "a", "Lake"};
char (*choice)[10];
if (flag == 1) {
    choice = &foo[0];
} else if (flag == 2) {
    choice = &bar[0];
} 
printf("%s%s\n", choice[0] , choice[1]);

For choice[1] to refer to the correct slot it has to be a pointer to the array element type and initialized to &foo[0] instead of &foo. Although they are the same address they are different data types.

If you want choice to be a pointer to the 2-dim char array, it can be done but you have specify both array dimensions when declaring the pointer and remember to dereference it:

char foo[][10] = {"Snakes", "on", "a", "Plane"};
char bar[][10] = {"Fishes", "in", "a", "Lake"};
char (*choice)[4][10];
if (flag == 1) {
    choice = &foo;
} else if (flag == 2) {
    choice = &bar;
} 
printf("%s%s\n", *choice[0] , *choice[1]);

How to select an array of strings in C?

2 Answers2