Why am I able to modify this const char* array?

Asked Apr 24 '19 at 16:02

Active Apr 24 '19 at 16:02

Viewed 59 times

Why does this work? It seems to me assigning a[0] = "X" should fail because const should make the data in the array immutable. However it compiles and runs fine, with and without const.

const char* a[] = {"a", "b", "c"};
a[0] = "X";
printf("%s\n", a[0]);

asked Apr 24 '19 at 16:02

Aviv Cohn

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What `a[i]` *points to* is immutable, but `a[i]` is not. `a[i]` is a *pointer to `const char`*. The pointer can be changed to point to a different `const char` object. – John Bode Apr 24 '19 at 16:04
it is : `declare a as array of pointer to const char` . So this one will not compile a[0][0] = 'x'; – 0___________ Apr 24 '19 at 16:09
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wrong dupe - it is something different – 0___________ Apr 24 '19 at 16:10

Why am I able to modify this const char* array?

0 Answers0