Thread: function calling: basic doubt

Hi,

I have a very basic doubt with calling a function,

i have a function that accepts a string of arrays and returns a string of array as below:

Code:

 char** parser(char **str)
    {
             return str;
           }

and i have the main calling it as :

int main()
{
    
      char label[50][25];

    
   	printf("enter expression");
	scanf("%s",inputexp);
	
    strcpy(label[0],inputexp);
            
	   
    char** test=parser(label);
     
    }

I am getting an error here.. can you please help me why.... i know it may be a very trivial case, but please do help me

If you get an error, surely you know what kind of error you are getting?

At base, though, I'm guessing it's because label isn't a char** and can't be made into one.

The error is:

cannot convert `char (*)[25]' to `char**' for argument `1' to `char** parser(char**)'

how do I set it right, i mean i need to assign the first element of string of array with a string of my choice and then pass that string of array into the function which outputs another string of array, please help

If you want to pass that array, then do so. Your function then needs to take char[][25] and can't take a char**, since that's not what an array is.

but the return type of the function cannot be an array right, i mean i need to return the array too, it can be done only with pointers right? correct me if iam wrong?

That is true, you cannot return an array. There is no way to make a pointer type out of a 2-D array, so either (1) you're not going to return an array -- pass it in instead or (2) you can't use "real" arrays, but instead you have to use "fake" arrays that you get from malloc.

str is not a **. If you want a 2-D array out of your function, pass it in and work with it:

Code:

void function_name(char in_array[][25], char out_array[][25])