string array initialisation

Question

This is a continuation of another question I have.

Consider the following code:

char *hi = "hello";

char *array1[3] = 
{
    hi,
    "world",
    "there."
};

It doesn't compile to my surprise (apparently I don't know C syntax as well as I thought) and generates the following error:

  error: initializer element is not constant

If I change the char* into char[] it compiles fine:

char hi[] = "hello";

char *array1[3] = 
{
    hi,
    "world",
    "there."
};

Can somebody explain to me why?

icabod · Accepted Answer · 2011-10-20 10:44:05Z

6

In the first example (char *hi = "hello";), you are creating a non-const pointer which is initialized to point to the static const string "hello". This pointer could, in theory, point at anything you like.

In the second example (char hi[] = "hello";) you are specifically defining an array, not a pointer, so the address it references is non-modifiable. Note that an array can be thought of as a non-modifiable pointer to a specific block of memory.

Your first example actually compiles without issue in C++ (my compiler, at least).

edited Oct 20, 2011 at 10:44

answered Oct 20, 2011 at 10:37

icabod

7,08427 silver badges42 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

lang2 Over a year ago

thanks. is there a way to use the const keyword to make the 1st piece of code work?

cnicutar Over a year ago

@lang2 Not in that scope. See this C FAQ. Not even with char *const hi (constant pointer to char).

icabod Over a year ago

Just added something, tho' I'm not sure if it will work... your first version compiles OK in C++ - I'll try with a C compiler when I get a moment :)

cnicutar Over a year ago

@icabod It won't. Read the link I posted.

Collectives™ on Stack Overflow

string array initialisation

1 Answer 1

4 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related