Writing an address to an array of pointers

Question

char *pointers_on_words[20]; char str[]= "First Second Third"; string_split(str,pointers_on_words); int string_split(char* string,char * array_results){ static int word_count; if( *string == '\0'){ int k= word_count; word_count =0; return k+1; }//конец строки if( *(string-1) == '\0'){// если предыдущий символ \0 - значит началось новое слово *array_results = string;//запись адреса нового слова array_results++; word_count++; } if(*string == ' '){//заменяем пробелы на \0 *string = '\0'; } return string_split(string+1,array_results); }

The result of this function should be a string broken into words separated by a \ 0 -character. And an array of pointers array_results, in which there should be pointers to the beginning of these very words.

But an error occurs when writing the address to the array of pointers.

I already gave you a ready-made code, that you again have some kind of nonsense?

Vlad from Moscow Vlad from Moscow one · Accepted Answer · 2016-11-28T17:15:56

You have declared the second parameter as char *

 int string_split(char* string,char * array_results){ ^^^^^^^^^^^^^^^^^^^^

so at least in this sentence

 *array_results = string;

a compilation error should occur since the left assignment operand is of type char , and the right operand is of type char * .

But array_results is after all an array of pointers, how else can it write an address?
@ Hardc0re you declared the parameter not as an array of pointers, but as a pointer.

Sergey Sergey 6,477 5 silver marks 19 bronze marks · Answer 2 · 2016-11-29T07:17:45

The second parameter of the function should be described like this:

 char * array_results[]

Accordingly, the code under the condition rewrite so:

 array_results[word_count] = string; word_count++;

What, in fact, will happen when a recursive function call with a static variable, I can not imagine ... Why is this necessary? After all, everything is solved by the method of linear viewing the line in one pass.

Sergey Gernichenko Sergey Gernichenko 24 6 bronze marks · Answer 3 · 2016-11-28T17:14:21

This is not a mistake, but a warning. You need to do a type conversion

Writing an address to an array of pointers

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