Say I want to dynamically allocate an array that can hold any data, i.e. using void**
as my type, in C. Then, rather than re-writing the pointer arithmetic logic each time, I want a simple function that returns the address of an element. Why is it that I cannot use this function in an assignment (i.e., so I can set, as well as get the element's value)?
Here's some example code:
#include <stdio.h>
#include <stdlib.h>
void printIntArray(void** array, size_t length) {
printf("Array at %p\n", array);
while (length--) {
printf(" [%zu] at %p -> %p", length, array + length, *(array + length));
if (*(array + length)) {
printf(" -> %d", *(int*)*(array + length));
}
printf("\n");
}
}
void* getElement(void** array, size_t index) {
return *(array + index);
}
int main(int argc, char** argv) {
const size_t n = 5;
size_t i;
/* n element array */
void** test = malloc(sizeof(void*) * n);
i = n;
while (i--) {
*(test + i) = NULL;
}
/* Set element [1] */
int testData = 123;
printf("testData at %p -> %d\n", &testData, testData);
*(test + 1) = (void*)&testData;
printIntArray(test, n);
/* Prints 123, as expected */
printf("Array[1] = %d\n", *(int*)getElement(test, 1));
/* Something new in [2] */
/* FIXME lvalue required as left operand of assignment */
int testThing = 456;
getElement(test, 2) = (void*)&testThing;
printIntArray(test, n);
return 0;
}
It wouldn't be the fist time this question has been asked, but the answer is usually along the lines of "you are trying to assign something to a function, rather than the return value of a function, which is not allowed". Fair enough, but can one get around this? I'm somewhat confused!!
Implement getElement()
like this:
void ** getElement(void ** ppv, size_t index) {
return ppv + index;
}
And use it like this:
*getElement(test, 2) = &testThing;
If you'd use a macro you could even go the way you intended:
#define GET_ELEMENT(ppv, index) \
(*(ppv + index))
Use it like this:
GET_ELEMENT(test, 2) = &testThing;