Question

void decimal2binary(char *decimal, char *binary) {   
    //method information goes here    
}

This is the main

int main(int argc, char **argv) {

char *data[100];
if (argc != 4) {
    printf("invalid number of arguments\n");
    return 1;
}
if (strcmp(argv[1] , "-d")) {

    if (strcmp(argv[3] , "-b")) {
        decimal2binary(temp, data);
    }
    }
}

Now I get this error

     warning: passing argument 2 of ‘decimal2binary’ from incompatible pointer type [enabled by default]

     note: expected ‘char *’ but argument is of type ‘char **’

So it says they are incompatible types but I have to use argv to get the data (that how I was asked) is there any other way?

Answer 1

Change the declaration of data to simply:

char data[100];

You don’t need an array of pointers to type char, which is what you’ve declared as your code stands right now. You simply want a byte array. I believe your confusion stems from the fact that while arrays are not pointers, they do decay into pointers to the first element of the array when passed as a function argument. So by simply saying decimal2binary(temp, data);, you are passing a pointer to the first element of data, and in this case you need that to be a pointer to a char, not a char*.