How to initialize char ** in C?

user_13830689 0
Viewed 581

I'm very new to C.

I'm supposed to do a simple wordsearch puzzle, so for the "dictionary" I did:

char **dictionary = {"DOG", "ELEPHANT", "CAT", "ETC", ""};

But when I try to compile, I get a warning saying 'excess elements in scalar initializer' for every word in the dictionary. Is the char ** wrongly initialized? How should I do it?

EDIT: My functions recieve char **dictionary.

Thanks in advance.

pointers char c string
4 Answers
const char *data[] = {"DOG", "ELEPHANT", "CAT", "ETC", ""};
const char **dictionary = data;
user_1407500 
char **values = (char *[]){"a", "b", "c"};

or

const char *array[] = {
    "First entry",
    "Second entry",
    "Third entry",
};
user_29772560 
const char **addresses = {"0xabcdefgh"};

produces only a warning

Incompatible pointer types initializing 'const char **' with an expression of type 'char [11]'
[clang -Wincompatible-pointer-types]

and it fails when accessing addresses[0].

Example:


/*
clang t.c && ./a.out
gcc t.c && ./a.out
 */
#include <stdio.h>
void printaddress0(const char **addresses) {
    printf(addresses[0]);
}
int main() {
    //const char **addresses = {"0xabcdefgh"}; // segmentation fault
    const char *addresses[] = {"0xabcdefgh"};
    printaddress0(addresses);
}
user_18038970 
//The best is to do like that
char **dictionary = new char*[] {(char*)"DOG", (char*)"ELEPHANT", (char*)"CAT", (char*)"ETC", (char*)""};

char* myword=new char[]{"my word"};
user_111908410
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