#include <stdio.h>
int main()
{
const int marbles[10] = { 1,2,3,4,5,6,7,8,9,10 };
int *ptr = marbles;
*ptr = 100;
printf("%d \n", marbles[0]); // output is "100"
}
I used const to protect the array. So I thought int *ptr = marbles would cause error. Otherwise by using pointer, it would enable user to change the data in the array. But surprisingly the output is "100". Isn't C suppose to protect the array from any kind of methods when I use const?
When I tried to compile the program, it showed this warning. So I guess when you initialize the pointer, const keyword loses its effect.
In function ‘main’:
try.c:6:16: warning: **initialization discards ‘const’ qualifier from pointer** target type [enabled by default]
int *ptr = marbles;
So I thought
int *ptr = marbleswould cause error.
It is an "error". C language considers this initialization a constraint violation, which is what we typically call an "error".
The compiler is required to issue a diagnostic message in response to a constraint violation, which is exactly what it did in your case.
Typically, it is your responsibility to analyze diagnostic message issued by your compiler and figure out which are "innocent" and which indicate serious errors.
