Pointer Pointing to Pointer

Question

Lets say I have a variable:

char** code;

I then do:

*code[0] = "Lucas"

Is it valid to say that, **code holds an array of pointers (*code is the array which I am making strings) and that *code[0] will equal "Lucas" and that *code[0][2] will equal 'c'?

Sorry if it seems elementary, I am getting very confused with double pointers! Thanks in advance!

-Lucas Giancola

You need to reserve space using malloc before, e.g: char **code = malloc(sizeof(*code) * n); and then code[0] = "Lucas" (without the *) — David Ranieri
– David Ranieri, Commented Feb 10, 2015 at 23:06
@AlterMann, You would probably get a warning/error for that because you'll be dropping the const from the string literal and will not have an array you can manipulate. — Brian McFarland
– Brian McFarland, Commented Feb 10, 2015 at 23:09
@BrianMcFarland, "Lucas" is a string literal (in a read-only segment), true, but there is nothing wrong in the assignment. — David Ranieri
– David Ranieri, Commented Feb 10, 2015 at 23:22
@BrianMcFarland: Not in C. String literals are of the type char* (actually char[N]), even though they're technically const char[]. This is an exception for mostly historical reasons, since C didn't always have const; it probably remains because a lot of existing code uses char* instead of const char*. — Tim Čas
– Tim Čas, Commented Feb 10, 2015 at 23:24

Iharob Al Asimi · Accepted Answer · 2015-02-10 23:15:44Z

3

Is it valid to say that, **code holds an array of pointers

No it is not. You have allocated space for a single pointer. You have told the compiler that that pointer will point at another pointer (that you have not created yet) that points to a character (that you have not created yet either)

*code[0] = "Lucas"

Is not valid code and doesnt compile

prog.cpp:6:8: error: invalid conversion from 'const char*' to 'char' [-fpermissive]
  *f[0] = "Lucas";

edited Feb 10, 2015 at 23:15

Iharob Al Asimi

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answered Feb 10, 2015 at 23:09

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Comments

R Sahu · Accepted Answer · 2015-02-10 23:23:43Z

2

If you just have

char** code;
*code[0] = "Lucas";

you'll run into undefined behavior since you did not allocate any memory for code.

You'll need to use:

char** code = malloc(SOME_SIZE*sizeof(*code));

Even after that, using:

*code[0] = "Lucas";

is not good. There are couple of problems with that.

code[0] is a pointer. *code[0] is that pointer dereferenced. If you already have some string in code[0], then, *code[0] will be first character of that string. The assignment is, therefore, wrong.
Also, "Lucas" is going to be in a read-only parts of the compiled code. You will need to make a copy of "Lucas" using strdup before you assign it to a variable that is of type char *.

You need something like:

 code[0] = strdup("Lucas");

Now you can use code[0] to access the whole string. You can use *code[0] to access the first character of that string. You can also use code[0][0] to access the first character of that string.

edited Feb 10, 2015 at 23:23

answered Feb 10, 2015 at 23:09

R Sahu

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1 Comment

Tim Čas Over a year ago

Warning: strdup is not standard C. It is a POSIX extension. You can easily write one yourself, though, via the use of malloc, strlen and memcpy --- just remember to name it differently (in case you are in a POSIX system).

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