Thread: how to create a function return with char* array

i am a beginner to use c programming.
how i am facing a problem about how to create a function with char * as a input parameter, and char *[] array as output parameter.

any code reference.

my steps:

1.input parameter char * = "abc/def/ghi";
2.in the function i want to slipt the char * into char * [] array by "/" as 3.delimiter. and finally out the char *[] as output.

now i have the code in main fuction but i want to create indivdutal function to handle slipt string.

Code:

int main ()
{
  char buf[] ="abc/qwe/ccd";
  int i;
  char *p;
  char *array[3];
  i = 0;
  p = strtok (buf,"/");  
  while (p != NULL)
  {
    array[i++] = p;
    p = strtok (NULL, "/");
  }
  for (i=0;i<3; ++i) 
    printf("%s\n", array[i]);
  return 0;
}

how to create a individual function to handle slipt string and output char * array[]?

Originally Posted by laserlight

It looks like you already did the hard part of implementing the splitting logic, so you just need to move the code over to a function that might be declared as:

Code:

void split(char buf[], char *array[]);

That said, you probably should have another parameter to specify the maximum number of elements of the array so as to avoid buffer overflow. The function could say, return the number of elements actually used instead of having void return type.

i want to create char*[] array as return object for split function, how can i do that?

Originally Posted by laserlight

You cannot actually return an array from a function, but you can return a pointer, so:

Code:

char **split(char buf[]);

The problem is that you cannot (or rather should not) return a pointer to an element of a non-static local array, so you have two common choices:

• Return a pointer to the first element of a static local array. This works very well for your specific example, but in practice this means the function has to decide to an upper bound for the array size that may prove to be too small.
• Return a pointer to dynamically allocated memory. This generally works very well, except that it means the caller is responsible for freeing the memory.

Another possibility is to go with my original suggestion and just return a pointer anyway, i.e., return array:

Code:

char **split(char buf[], char *array[]);

but it may be a little pointless

i want to ask what is the different between char *[] and char**[]? if i return the pointer . how to use for loop to get all the elment of the returned array?

Originally Posted by laserlight

As used in the parameter list of a function, the former means "pointer to pointer to char" and the latter means "pointer to pointer to pointer to char". My guess is that you actually wanted to ask what char** meant in the return type, in which case it just means "pointer to pointer to char", but you should already know this.


You have already done such a loop to print in your post #1.

By the way, are you really trying to learn C and C++ at the same time? It is not impossible, especially if you already have a programming background, but for people new to programming it is a recipe for confusion and frustration.

what is former = char* < only 1 *
and latter char** ?
is it correct?