I have a question to the conversion of int array to char*. The following code has the output 23. But I don't really understand why. Can someone explain it to me?
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
int main(){
uint32_t x;
uint32_t* p = (uint32_t*) malloc(sizeof(uint32_t));
uint32_t array[9] = {42, 5, 23, 82, 127, 21, 324, 3, 8};
*p = *((char*)array+8);
printf("1: %d\n", *p);
return 0;
}
The size of a uint32 is 32 bits, or 4 bytes. When you do (char*)array+8, you cast the array into an array of char, and take the eighth character. Here, the eighth character contains the beginning of the integer 23, which fits into a char.
*p = *((char*)array+8*sizeof(uint32_t)
to move to array[8] in your example you are only moving 8 bytes forward
If you think about it, you have casted the array to char*. Where sizeof char is 1B.
Which means usage pointer arithmetics in this case +8 doesnt move you 8*sizeof(uint32_t) to nineth element of array, but only 8 Bytes (8*sizeof(char)).
Since uint32_t has 4 Bytes, you have moved to first byte of 3rd element 23.
First you have to use pointer arithmetics on uint32_t array and after that cast it to char, like
*p = (char)*(array+8); // Prints 8