c array concatenation problem

Question

I'm trying to concatenate an array with the array from argv (I forgot the formal name for that, input array? parameter array?)

anyway, I initalize the original array, then I use memcpy to copy the two arrays into the new array.

int main(int argc, char *argv[]) {
  char *args1[] = {"foo","bar"};
  char **args = (char**) calloc(argc, sizeof(char*));
  memcpy(args, args1, sizeof(char*) * 2);
  memcpy(args + sizeof(char*) * 2, argv+1, sizeof(char*) * (argc-1));

but when I run this

  printf("%s %s %s\n", args[0], args[1], args[2]);

and then run

./test baz

I get a result of

foo bar (null)

So I'm trying to figure out where my second memcpy screwed up, but I can't find it. Any help? Thanks in advance.

chrisaycock · Accepted Answer · 2011-01-05 22:20:38Z

5

There are two problems in this code.

Problem #1

calloc(argc, sizeof(char*))

should be

calloc(argc + 2, sizeof(char*))

Problem #2

You are confusing pointer arithmetic. This part

args + sizeof(char*) * 2

should simply be

args + 2

Alternatively, you could use

&args[2]

Either way, don't multiply the offset by the size of the type. The compiler does that automatically.

answered Jan 5, 2011 at 22:20

chrisaycock

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1 Comment

kei-clone Over a year ago

sigh. I know my pointer arithmetic, so it was silly of me to make that kind of error and not notice it. That's the last time I take pointers from rosettacode

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