What is the right way to initialize char** ?
I get coverity error - Uninitialized pointer read (UNINIT) when trying:
char **values = NULL;
or
char **values = { NULL };
3 Answers
This example program illustrates initialization of an array of C strings.
#include <stdio.h>
const char * array[] = {
"First entry",
"Second entry",
"Third entry",
};
#define n_array (sizeof (array) / sizeof (const char *))
int main ()
{
int i;
for (i = 0; i < n_array; i++) {
printf ("%d: %s\n", i, array[i]);
}
return 0;
}
It prints out the following:
0: First entry
1: Second entry
2: Third entry
2 Comments
andymcgregor
Hi, it works! But i have a question, why i have to use #define n_array here? I have tried with: int n_array, and my app crashes, i'm new to C and dont understand this behavior - #define instead of simple variable int. Thanks.
Duncan Jones
@teMkaa Using
#define is just a question of style; personally I wouldn't have done it. Within the main function you could easily define another int variable and assign it to sizeof (array) / sizeof (const char *).Its fine to just do char **strings;, char **strings = NULL, or char **strings = {NULL}
but to initialize it you'd have to use malloc:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(){
// allocate space for 5 pointers to strings
char **strings = (char**)malloc(5*sizeof(char*));
int i = 0;
//allocate space for each string
// here allocate 50 bytes, which is more than enough for the strings
for(i = 0; i < 5; i++){
printf("%d\n", i);
strings[i] = (char*)malloc(50*sizeof(char));
}
//assign them all something
sprintf(strings[0], "bird goes tweet");
sprintf(strings[1], "mouse goes squeak");
sprintf(strings[2], "cow goes moo");
sprintf(strings[3], "frog goes croak");
sprintf(strings[4], "what does the fox say?");
// Print it out
for(i = 0; i < 5; i++){
printf("Line #%d(length: %lu): %s\n", i, strlen(strings[i]),strings[i]);
}
//Free each string
for(i = 0; i < 5; i++){
free(strings[i]);
}
//finally release the first string
free(strings);
return 0;
}
2 Comments
V-X
(I'd expext some c-grammar nazzi commenting about the typecast before malloc)
Nick Beeuwsaert
I guess it's just habit from writing C++ and having to do that
There is no right way, but you can initialize an array of literals:
char **values = (char *[]){"a", "b", "c"};
or you can allocate each and initialize it:
char **values = malloc(sizeof(char*) * s);
for(...)
{
values[i] = malloc(sizeof(char) * l);
//or
values[i] = "hello";
}
3 Comments
Brian Tracy
using
malloc() is the way to go
Sebastian
I get compiler error
taking address of temporary array when trying to compile char **values = (char *[]){"a", "b", "c"}; ... char *values[] = {"a", "b", "c"}; works however
quant2016
I googled "c inline array of strings" and you line with
(char *[]){"a", "b", "c"} matches my request ideally.
char ** valuesis a pointer to a pointer tochar, nothing more.valuesis no array of "strings".char *values[] =to allocate the array on stack?