Reference to Array vs reference to array pointer

Question

void check(void* elemAddr){
    char* word = *((char**)elemAddr);
    printf("word is %s\n",word);
}

int main(){
    char array[10] = {'j','o','h','n'};
    char * bla = array;
    check(&bla);
    check(&array);
}

Output:

word is john

RUN FINISHED; Segmentation fault; core dumped;

First one works, but second not. I don't understand why this happens.

Answer 1

The problem is, when we do &array, we are getting a char (*)[10] from an char [10], instead of a char **.

Before we do our experiment, I will emphasize that, when we pass an array as an argument to a function, C actually casts the array to a pointer. The big bucket of data is not copied.

Thus, int main(int argc, char **argv) is identical to int main(int argc, char *argv[]) in C.

This made it available for us to print the address of an array with a simple printf.

Let's do the experiment:

char array[] = "john";
printf("array:  %p\n", array);
printf("&array: %p\n", &array);

// Output:
array:  0x7fff924eaae0
&array: 0x7fff924eaae0

After knowing this, let's dig into your code:

char array[10] = "john";
char *bla = array;
check(&bla);
check(&array);

bla is char *, and &bla is char **.

However, array is char [10], and &array is char (*)[10] instead of char **.

So when you pass &array as an argument, char (*)[10] acts like a char * when passing as an argument, as is said above.

Therefore **(char **) &bla == 'j' while *(char *) &array == 'j'. Do some simple experiments and you will prove it.

And you are casting void *elemAddr to a char ** and try to deference it. This will only work with &bla since it is char **. &array will cause a segfault because "john" is interpreted as an address as you do the cast.