14

I want a simple function that receives a string and returns an array of strings after some parsing. So, this is my function signature:

int parse(const char *foo, char **sep_foo, int *sep_foo_qty) {
    int i;
    char *token;
    ...
    strcpy(sep_foo[i], token); /* sf here */
    ...
}

Then I call it like this:

char sep_foo[MAX_QTY][MAX_STRING_LENGTH];
char foo[MAX_STRING_LENGTH];
int sep_foo_qty, error;

...

error = parse(foo, sep_foo, &sep_foo_qyt);

...

This way I get a warning during compilation:

warning: passing argument 2 of 'parse' from incompatible pointer type

And then a segmentation fault during execution in the line marked with /* sf here */

What is wrong in my C code?

Thanks in advance

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4 Answers 4

Reset to default
28

The warning is exactly right. Your function wants an array of pointers. You're giving it an array of arrays.

Expected:

 sep_foo:
 +------+       +-----+
 |char**|--> 0: |char*|-->"string1"
 +------+       +-----+
             1: |char*|-->"string2"
                +-----+
*sep_foo_qty-1: |...  |
                +-----+

What you provided:

           sep_foo:
           +--------------------------------+
        0: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
        1: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
MAX_QTY-1: | ...                            |
           +--------------------------------+

An array with elements of type X can "decay" into a pointer-to-X, or X*. But the value of X isn't allowed to change in that conversion. Only one decay operation is allowed. You'd need it to happen twice. In your case, X is array-of-MAX_STRING_LENGTH-chars. The function wants X to be pointer-to-char. Since those aren't the same, the compiler warns you. I'm a bit surprised it was just a warning since nothing good can come from what the compiler allowed to happen.

In your function, you could write this code:

char* y = NULL;
*sep_foo = y;

That's legal code since sep_foo is a char**, so *sep_foo is a char*, and so is y; you can assign them. But with what you tried to do, *sep_foo wouldn't really be a char*; it would be pointing to an array of char. Your code, in effect, would be attempting to do this:

char destination[MAX_STRING_LENGTH];
char* y = NULL;
destination = y;

You can't assign a pointer into an array, and so the compiler warns that the call is no good.

There are two ways to solve this:

  • Change the way you declare and allocate sep_foo on the calling side so it matches what the function expects to receive:

    char** sep_foo = calloc(MAX_QTY, sizeof(char*));
for (int i = 0; i < MAX_QTY; ++i)
  sep_foo[i] = malloc(MAX_STRING_LENGTH);

    or, equivalently

    char* sep_foo[MAX_QTY];
for (int i = 0; i < MAX_QTY; ++i)
  sep_foo[i] = malloc(MAX_STRING_LENGTH);

  • Change the prototype of the function to accept what you're really giving it:

    int parse(const char *foo, char sep_foo[MAX_QTY][MAX_STRING_LENGTH], int *sep_foo_qty);
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14 Comments

@che, An array can be converted to a pointer by taking the pointer of its first element (ptr = &array[0]). This can be done implicitly (ptr = array). The reverse is not true.
Also, when you take the pointer of an array (ptr = &ar), ptr contains the address of the array as a whole. ptr would then be equal to &ar[0], in general, except the type is different. If you took the pointer of a pointer, you would get a number which, is not equal to that pointer (memory is ref'd).
On your edit... You did a much better job at a diagram than I did. I'm the opposite of artistic. =]
Rob, i like you answer very much. i also like how that answer nicely complements my recent pet peeve: stackoverflow.com/questions/423823/… . I will put a link on that to your answer. have fun :)
Rob, I think you mistyped the variable name: sep_foo_qty is int*, sep_foo is char**. Please edit your answer because is really good but a bit confusing there.
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15

Parameter 2 should be

char sep_foo[][MAX_STRING_LENGTH]

To clarify, you are passing a pointer to parse() and treating it as a pointer to a pointer. A multidimensional array in C is NOT an array of pointers. It is a single block of memory that is pointed to by the array variable. You cannot dereference it twice.

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1 Comment

I've been trying to understand strager's memory models, but you've nailed the difference in three clear and simple sentences. Thanks for the clarification!
4

sep_foo is defined as an array of arrays. In other words, when you use sep_foo, it points to the beginning of sequential memory. Here's a model:

(assume MAX_STRING_LENGTH = 16, MAX_QTY = 2)
sep_foo       = &&0000
sep_foo[0]    =  &0000
sep_foo[0][0] = *&0000 = 12
sep_foo[0][8] = *&0008 = 74
sep_foo[1]    =  &0010
sep_foo[1][0] = *&0010 = 12


0000  12 34 56 78  9A BC DE F0  74 10 25 89  63 AC DB FE
0010  12 34 56 78  9A BC DE F0  74 10 25 89  63 AC DB FE

However, your function expects an array of pointers (actually, a pointer to a pointer). This is modeled as such:

sep_foo_arg       =   &&0000
sep_foo_arg[0]    =  *&&0000 = &0010
sep_foo_arg[0][0] =  *&*&0000 = 12
sep_foo_arg[0][8] = *(&*&0000 + 8) = 74
sep_foo_arg[1]    =  *&&0002 = &0020
sep_foo_arg[1][0] = *&*&0000 = 12

0000  0010 0020  xxxx xxxx  xxxx xxxx  xxxx xxxx

0010  12 34 56 78  9A BC DE F0  74 10 25 89  63 AC DB FE
0020  12 34 56 78  9A BC DE F0  74 10 25 89  63 AC DB FE

Yeah ... Syntax may be a bit confusing for my explanations...

Anyway, you can solve this issue by telling your function how to treat the pointer pointed to. In particular, you would want to treat it as an array (a sequence of memory):

int parse(const char *foo, char (*sep_foo)[MAX_STRING_LENGTH], int *sep_foo_qty);
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9 Comments

strager, didn't you mean char (sep_foo)[MAX_STRING_LENGTH] ? because as it is now, it's still a "char*". looks like a typo to me.
@litb, Ah, right. Normally I look this up, but I was in a hurry when I wrote it (couldn't you tell? =]). Thanks.
@litb, Wait -- did YOU make a typo? I don't know which one is right any more. @_@ Some testing is in order later.
i'm sorry i see i sounded confusing. well i mean you made teh typo :) look: T f[N] is an array of T. T (*f)[N] is a pointer to an array of T (what you wanted). the * binds more to the T than to the parameter name, so you have to paren it to the f, so that f becomes a pointer.
because if you say "T *f[N]", it is the same thing as "T **f" in the parameter list. i don't believe you meant that.
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-2

If that is your exact code, then I'm guessing the segfault is because of the fact that you haven't allocated memory to the char* token inside your parse function, and then using that in your strcpy.

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1 Comment

I don't need to allocate memroy for char *token as I'm using as the result of strtok_r function: token = strtok_r(copied_foo, "/", &save_ptr); ... token = strtok_r(NULL, "/", &save_ptr); Thanks

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