Checking if my array of pointer string is an integer-CodePudding

EDIT -> My problem with finding if was an integer got solved. My next problem is how to get the size of the array. Example:

char **str = [hello][world] -> returns 2

char **str = [hello] -> return 1

char **str = [hello][world][yes] -> returns 3

The way I am doing right now. It's wrong

int size = sizeof(str) / sizeof(str[0]);
I also tried
int size = sizeof(str) / sizeof(char *);

.....

I have a program that gets a line from the user like "Hello World". Then I parse that into an array of pointers: so it's something like char **str = [Hello][World].

Then I want to check if the second index is an integer or not. So if the user types Hello 2, this function returns true (1), but if not returns false(0).

I am a little confused how to work with pointers.

So I think I will have something like this:

int my_function(char** str) {
    // it's empty
    if (str[0] == NULL)
        return 0;

    // Just have one index
    int size = sizeof(str) / sizeof(str[0]);
    if (size < 1) {
        return 0;
    }

    // checking if data in second index is an integer
    int length = strlen(str[1]);
  
    for (int i = 0; i < length; i  ) {
        // Here is my problem how do I go over each character for 
        // only str[1] to check if is an integer???????
        if (!isdigit(str)) {
            return 0;
        }
    }

    return 1;
}

CodePudding user response：

You need a double dereference. The first dereference to get the relevant char pointer. The second dereference to get the relevant character.

Try:

isdigit(str) -->  isdigit(str[1][i])

CodePudding user response：

Try following,

    char *secondString = str[1];
    int length = strlen(secondString );
    
    for (int i = 0; i < length ;   i)
    {
    if (!isdigit(secondString[i]))
    return 0;
    }

return 1;