I am just trying to write something like that:
u64 get_env(char *argv[]);
char* g_argv[];
static char * Sample ()
{
return (char*)get_env(g_argv);
}
int main(int argc, char *argv[])
{
g_argv = argv;
Sample();
}
Getting error: 'g_argv' has an incomplete type warning: array 'g_argv' assumed to have one element [enabled by default]
I've tried many different ways. How to write it right?
int main(int argc, char *argv[])
Despite what it looks like, argv is not an array; arrays aren't first class objects in C and cannot be passed to functions, only their addresses can. So because argv is a parameter, it's actually a pointer to a pointer. For this reason I think it's better to tell the truth and use
int main(int argc, char** argv)
which is exactly equivalent to the above. This confusion in the language has led you to
char* g_argv[];
You're saying this is an array of pointers, without saying how big the array is, but that's not what you want; you want a pointer to the first of several pointers:
char** g_argv;
That fixes the problem you asked about, but I wonder about this declaration:
u64 get_env(char *argv[]);
Why declare it as returning u64 when the name and usage clearly indicate that it returns a char*? Actually, you should not be declaring it here at all ... it should be declared in a header file that specifies the API that includes get_env. Hopefully that header file declares it as returning a char*, and then you can remove the cast from
return (char*)get_env(g_argv);
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