How to print the contents of char**?

Question

I have a structure defined as a char** array containing strings. I don't know how to run printf on its contents.

#include <stdio.h> #include <string.h> #include <stdlib.h> #ifndef STRUCT_STRING_ARRAY #define STRUCT_STRING_ARRAY typedef struct s_string_array { int size; char** array; } string_array; #endif void my_print_words_array(string_array* param_1) { int len = param_1->size; char **d = param_1->array; for(int i = 0 ; i < len;i++){ printf("%s\n", d[i]); } } int main(){ struct s_string_array *d; d->size = 2; char **my_arr = (char *[]){"hello", "world"};//this init is fine d->array = my_arr; my_print_words_array(d); return 0 ; }

The main function gives me segfault error. What's wrong?

@RobertHarvey hi , u mean it works? no repro? you mean i should just put a link to executable online ide? — ERJAN
– ERJAN, Commented May 21, 2020 at 17:18
Well... I'm inclined to agree with Tony. You're not allocating memory for struct s_string_array *d; — Robert Harvey
– Robert Harvey, Commented May 21, 2020 at 17:19

Vlad from Moscow · Accepted Answer · 2020-05-21 17:31:39Z

There is no sense to declare a pointer to the structure

struct s_string_array *d;

moreover that is not initialized and has indeterminate value that further is a reason of undefined behavior.

What you are trying to achieve is the following

#include <stdio.h> typedef struct s_string_array { int size; char** array; } string_array; void my_print_words_array( const string_array *param_1 ) { for ( int i = 0; i < param_1->size; i++ ) { puts( param_1->array[i] ); } } int main( void ) { string_array d = { .size = 2, .array = (char *[]){"hello", "world"} }; my_print_words_array( &d ); return 0 ; }

The program output is

hello world

Good example of using char ** without using dynamic allocated memory.

Collectives™ on Stack Overflow

How to print the contents of char**?

1 Answer 1

1 Comment

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Related