Question

In C++ I can allocate an array of strings (say 10 strings) very easily as:

string* arrayOfString = new string[10];

however I don’t know how to do it in ANSI C. I tried:

char** arrayOfString = (*(char[1000])) malloc (sizeof (*(char[1000])));

But the compiler (MinGW 3.4.5) keeps saying that it is “Syntax error”.
How to do it right?
Thanks.

Answer 1

If each string is of same size which is known at compile time, say it is 100, then you can do this¹:

typedef char cstring[100]; //cstring is a new type which is 
                           //a char array of size 100

cstring *arrstring = malloc (1000 * sizeof (cstring));

int i;
for( i = 0 ; i < 1000 ; ++i)
     strcpy(arrstring[i], "some string of size less than or equal to 100");

for( i = 0 ; i < 1000 ; ++i)
     printf("%s\n", arrstring[i]);

Demo : http://ideone.com/oNA30

^{1. Note that as @Eregrith pointed out in the comment that cast is not advised if you compile your code as C. However, if you compile it as C++, then you need to write (cstring*)malloc (1000 * sizeof (cstring)) but then in C++, you should avoid writing such code in the first place. A better alternative in C++ is, std::vector<std::string> as explained at the bottom of this post.}

If the size of each string is not known at compile time, or each string is not of same size, then you can do this:

char **arrstring =  malloc(sizeof(char*) * 1000); //1000 strings!
int i;
for(i = 0 ; i < 1000; ++i)
     arrstring[i] = (char*) malloc(sizeof(char) * sizeOfString); 

I’m assuming same size sizeOfString for all 1000 strings. If they’re different size, then you’ve to pass different value in each iteration, something like this:

for(i = 0 ; i < 1000; ++i)
     arrstring[i] = malloc(sizeof(char) * sizeOfEachString[i]); 

I hope that helps you, and I also hope you can do the rest yourself.

---

By the way, in C++, you should not do this:

string* arrayOfString = new string[10]; //avoid new as much as possible!

Instead, you should do this:

std::vector<std::string>  strings;
strings.reserve(10);