How to find number of rows in dynamic 2D char array in C?

问题

I have an input function like :

int func(char* s[])
{
    // return number of rows in the character array s
}

where the char array s is provided randomly as for eg: {"sdadd", "dsdsd", "dsffsf", "ffsffsf"}.

Here output should be 4 for above example.

回答1:

Passing string arrays requires more information...

This is similar to the input you would find in say a call to int main(int argc, char *argv[]). The exception is that when main() is called, it reads the command line and provides count information, in argc, to determine the number of arguments in argv[]. That is why the char *argv[] argument can be passed.

You have the same requirement. That is when passing an argument such as char *s[], i.e., you must also somehow provide a value telling func() how many strings. C has no way of knowing the number of strings in that variable without being told. This is because an array reference ( char *s[]; ) decays into a pointer to char ( char *s; ) pointing to the first element of that array, i.e., no array size information.

So, the answer to your question is:, with the information given, func() cannot determine the number of strings in s.

An important distinction:
The size CAN be determined for character arrays, such as

 char *s[]={"this","is","an","array"};  // the assignment of values in {...}
                                        //make this an array of strings.

But only when s is declared in the same scope of the function attempting to determine the number of elements. If that condition is met, then use:

int size = sizeof(s)/sizeof(s[0]);  //only works for char arrays, not char *  
                                    //and only when declared and defined in scope of the call

In scope, in this context, simply means that char *s[]... be defined with global visibility, or within the function calculating number of elements. If passed as an argument, char *s[];, as defined above, will be simply seen as char *s, and the number of arguments cannot be determined.

Other options for retrieving count of strings in a string array include:
1) Modify your input array so that something in the last string is unique, like "%", or "\0". This will allow you to test strings with standard C string functions.

2) Include an argument in the function providing number of strings. (eg, similar to main(...) or printf(...))

回答2:

Like with all array parameters in C there is no way to do that, because arrays decay to a pointer to their first element (for a discussion cf. the C FAQ, http://c-faq.com/aryptr/) when passed as function parameters.

As always in this case you have two possibilities.

Use an end marker. That works always if the element type of the array can hold a special value which does not occur "naturally" for that type. For chars that's '\0', for pointers like in your array that's NULL, the null pointer. For example argv[], the second argument to main(), is terminated by a null pointer. That makes iterating easy:
```
#include<stdio.h>
int main(int argc, char *argv[])
{
    int i;
    for(i=0; argv[i] != NULL; i++)
    {
        printf("%s\n", argv[i]);
    }
    return 0;
}
```
Pass a length information with the array. That's, interestingly, the information in argc which the C runtime also passes to main().

Beyond that there is no (or at least no portable) way to know how many elements an array has which is passed as a parameter to a function in C.

回答3:

int size = sizeof(s)/sizeof(s[0]);

This statement will do. The variable size will be assigned the number of strings present in the *s[].

来源：https://stackoverflow.com/questions/19481251/how-to-find-number-of-rows-in-dynamic-2d-char-array-in-c

标签

multidimensional-array

arrays