dereferencing pointer to incomplete type

[Steven]

Hello,

I get the following message during compilation:
`dereferencing pointer to incomplete type'

It stems from the compare function I use with qsort() and I am not
quite sure how to fix it.

I have a struct like:
struct node {
time_t utc;
};

And the code that generates the error during compilation:
int cmputc(const void *x, const void *y) {
struct tnode * const *a = x;
struct tnode * const *b = y;
int retv = 0;

if((*a)->utc < (*b)->utc)
retv = -1;
else if((*a)->utc == (*b)->utc)
retv = 0;
else
retv = 1;

return retv;
}

Does anyone know what I am doing wrong ?

Thank you for your explanation.

Steven.

 

[Steven]

Ooopss..
Sorry. already solved..... It's in the name using node in the
structure declaration and tnode in the compare function .. aiiii..

Thankx anyway.. [blush]

 

[Emmanuel Delahaye]

Steven a écrit :
Be sure that this structure definition ...

struct node {
time_t utc;
};

.... is visible from ...

int cmputc(const void *x, const void *y) {
struct tnode * const *a = x;
struct tnode * const *b = y;

.... this function. Typically, it sould be placed in a header and
included where you need it.

 

[Christopher Benson-Manica]

I have a struct like:
struct node {
time_t utc;
};
Right...

int cmputc(const void *x, const void *y) {
struct tnode * const *a = x;
struct tnode * const *b = y;

So what's a struct tnode? Perhaps this is related to your problem?

 

[dwks]

int cmputc(const void *x, const void *y) {

struct tnode * const *a = x;
struct tnode * const *b = y;

You can't dereference a void pointer; you need to cast it.

int cmputc(const void *x, const void *y) {
struct tnode * const *a = (struct tnode * const *)x;
struct tnode * const *b = (struct tnode * const *)y;

 

[Flash Gordon]

dwks wrote:

Please leave in attributions so we can see who you are replying to.

You can't dereference a void pointer; you need to cast it.

The above is not dereferencing anything, it is assigning and is
perfectly valid in C. It is C++ which does not allow it, but that is a
different language.

int cmputc(const void *x, const void *y) {
struct tnode * const *a = (struct tnode * const *)x;
struct tnode * const *b = (struct tnode * const *)y;

Your suggestion above is far harder to read, so I would recommend
sticking with the perfectly legal option of not casting.