warning with const operator (in struct) - c

Question

it may be a nooby question but i cant figure out why i get a warning
struct4.c:32:15: warning: assignment discards ‘const’ qualifier from pointer target type [enabled by default] crea[i].size = wsize[i%5];
compiling this:
struct shirt {
char *size;
char *colour;
} ;
typedef struct shirt Camicia;
void loadshirt (Camicia * const crea, const char *wsize[] , const char *wcolour[]);
int main (void) {
Camicia collezione[50];
const char *sizearray[] = {"xs","s","m","l","xl"};
const char *colourarray[] = {"black","blue","yellow","orange"};
loadshirt(collezione,sizearray,colourarray);
printf("%s\n",collezione[4].size);
printf("%s\n",collezione[4].colour);
return 0;
}
void loadshirt (Camicia * const crea, const char *wsize[] , const char *wcolour[]) {
int i=0;
while (i<50) {
crea[i].size = wsize[i%5];
crea[i].colour = wcolour[i%4];
i++;
}
}

Answer 1

You defined data members of the structure as pointers to non-const character strings.
struct shirt {
char *size;
char *colour;
} ;
However in the function you are assigning pointers to const character strings to pointers to non-const character strings
crea[i].size = wsize[i%5];
crea[i].colour = wcolour[i%4];
See declarations of wsize and wcolour in the parameter list
const char *wsize[] , const char *wcolour[]
You may not do that.
Define the data members as pointers to const strings
struct shirt {
const char *size;
const char *colour;
} ;
Or define the parameters as having type of pointers to non-const strings
char *wsize[] , char *wcolour[]
In this case you have to change also the definitions of the corresponding arguments
char *sizearray[] = {"xs","s","m","l","xl"};
char *colourarray[] = {"black","blue","yellow","orange"};
In C strings literals have types of non-const arrays.

Answer 2

Here your function takes a const pointer to a Camicia
void loadshirt (Camicia const * crea, const char *wsize[] , const char *wcolour[]) {
And 3 lines later you try to modify crea :
crea[i].size = wsize[i%5];
crea[i].colour = wcolour[i%4];
You can't do that.
When the compiler says something along the lines of X discards 'const' qualifier it means exactly that. Something is const, but you're trying to modify it as if it weren't.
It's important to try to understand the compiler's error messages, you'll save a lot of time.
Now if you want to fix the function, first you'll need to remove the const qualifier from crea in the list of parameters.
But also note that here wsize and wcolour are const, while Camicia is defined like this:
struct shirt {
char *size;
char *colour;
} ;
typedef struct shirt Camicia;
Either make your struct Camicia store const char* or modify the other parameters to be char*. Since you're using string literals in main, you'll probably want everything to be const char*.

Answer 3

to avoid the runtime seg fault events,
the following code will work correctly.
this code takes into account that the arrays in main()
are actually an array of pointers to char* (I.E. strings)
#include <stdio.h>
#include <stdlib.h>
#define MAX_SHIRTS (50)
struct shirt {
const char *size;
const char *colour;
} ;
// struct shirt * const says the pointer is const,
// not that the contents of where the pointer points is const
void loadshirt (struct shirt * const, const char **, const char **);
int main (void) {
struct shirt collezione[MAX_SHIRTS];
// create two arrays of const pointers to consts
const char const *pSize[] = {"xs","s","m","l","xl"};
const char const *pColour[] = {"black","blue","yellow","orange"};
loadshirt(collezione, pSize, pColour);
printf("%s\n",collezione[4].size);
printf("%s\n",collezione[4].colour);
return 0;
}
void loadshirt (struct shirt * const crea, const char **pSize , const char **pColour)
{
int i=0;
for(i=0; i<MAX_SHIRTS; i++)
{
crea[i].size = pSize[i%5];
crea[i].colour = pColour[i%4];
}
}

Answer 4

You are making a non-const pointer to a const pointer, you can't modify the contents the const pointer points to, but if you discard const qualifier you can do it through the new non-const pointer, so the compiler is warning you about that.