int const array cannot be written to

Question

Compiling the following program

int main(void) {
    int const c[2];
    c[0] = 0;
    c[1] = 1;
}

leads to error: assignment of read-only location ‘c[0]’. As I understand it, the const only applies to the location of c and so c[0] and c[1] should be mutable. Why is this error produced?

"int const array cannot be written to" Indeed, that is the purpose of const. — Lundin
– Lundin, Commented Jan 13, 2020 at 14:25
Oh btw, ` int const c[2];` with semicolon and no initializer list is kind of a language bug. The C language allows you to declare a constant without initializing it. C++ does not. — Lundin
– Lundin, Commented Jan 13, 2020 at 14:29
To get the behavior you describe in the question you could write int (*const c)[10]; *c[1] = 5; — xate
– xate, Commented Jan 13, 2020 at 14:29

Blaze · Accepted Answer · 2020-01-13 14:25:48Z

8

As I understand it, the const only applies to the location of c

No. You can't modify the location of the array anyway. What you probably mean is if you have a int * const, then that indeed is a constant pointer to a modifiable int. However, int const c[2]; is an array of 2 constant ints. As such, you have to initialize them when you declare the array:

int const c[2] = {0, 1};

In constrast:

int main(void) {
    int c[2];
    int* const foo = c;
    foo[0] = 0;
    foo[0] = 1;
    //foo = malloc(sizeof(int)); doesn't work, can't modify foo, as it's constant
}

edited Jan 13, 2020 at 14:25

answered Jan 13, 2020 at 14:22

Blaze

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6 Comments

Andreas Wenzel Over a year ago

It doesn't matter if you write "int const" or "const int", does it? I have never seen "int const" written before. (I am aware that this was taken over from the question.)

0___________ Over a year ago

the order does not matter in this case.

Lundin Over a year ago

@AndreasWenzel const int and int const are equivalent - the latter is weird style. const int* and int * const however, mean different things.

Lundin Over a year ago

(We can even write const const const int x. Because.... C...)

Lundin Over a year ago

@Blaze All that matters is if it's on the left or right side of the *. Your "left of it" rule doesn't hold. Consider auto const const int const signed const x = 0;

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