Re-Initializing a const char array in C is not giving error

Question

While initializing a const char array, I tried to change the string and I was able to change it without any issue.

I was learning how to initialize a const char array. I think I am doing some mistake here which I am not able to find.

int main(int argc, char const *argv[])
{

    const char *strs[10];
    strs[0] = "wwww.google.com";
    printf("%s\n", strs[0]);
    strs[1] = "https://wwww.google.com";
    strs[0] = "ss";
    printf("%s\n", strs[0]);
    return 0;
}

Output:

1st init: wwww.google.com
2nd init: ss -> Here, I expect it to throw error

The pointers you’re setting (not initializing) are not constant; the data they point at can’t be modified via the pointers, though. You don’t try to modify the strings; all is well. — Jonathan Leffler
– Jonathan Leffler, Commented Jul 14, 2019 at 7:33
If the pointers were constant, your assignments would generate compiler warnings. — Jonathan Leffler
– Jonathan Leffler, Commented Jul 14, 2019 at 7:36
1st init. Nah, the pointers are default initialized first. And then you assign for a second time with "wwww.google.com". — Ivaylo Valchev
– Ivaylo Valchev, Commented Jul 14, 2019 at 7:38
const char *strs[10]; defines strs as an array of ten pointers to constant characters. The array itself isn't constant, and neither is it an array of characters. — Some programmer dude
– Some programmer dude, Commented Jul 14, 2019 at 7:39
Incidentally, those are assignments, not reinitializations. Initialization is something that happens when an object is created, and you cannot do it more than once. — Eric Postpischil
– Eric Postpischil, Commented Jul 14, 2019 at 11:24

Spikatrix · Accepted Answer · 2019-07-14 07:36:52Z

3

const char* s = "Hi";

tells the compiler that the content that the pointer points to is constant. This means that s[0] = 'P'; will result in a compilation error. But you can modify the pointer. On the other hand,

char* const s = "Hi";

tells the compiler that that the pointer is constant. This means that s = "Pi"; will result in a compilation error. But no compilation error will be thrown when you try to modify the string^*

Your code depicts the former behaviour, not the latter as you seem to have thought

^{* Modifying string literals will invoke Undefined Behaviour}

answered Jul 14, 2019 at 7:36

Spikatrix

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pmg · Accepted Answer · 2019-07-14 08:32:07Z

2

const char *strs[10];

strs is an array of 10 pointers to const char. You can change the pointers; you cannot change the chars

strs[2] = NULL; // ok: change the pointer
strs[0][0] = '#'; // wrong; cannot change the char

Maybe try

const char * const strs[10] = {"www.google.com",
                               "https://www.google.com",
                               "www.google.com/",
                               "https://www.google.com/",
                               NULL, NULL };

which makes strs an array of 10 read-only pointers to const char. You cannot change the pointers after initialization.

answered Jul 14, 2019 at 8:32

pmg

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Tsepo Nkalai · Accepted Answer · 2019-07-14 07:46:51Z

0

To put this in simple English(not necessarily 100% accurate but serves to conceptualise), this

const char *strs[10];

initialises a contant array strs which contains none constant elements. Thus, the elements in the array can be changed but the array itself cannot be changed

answered Jul 14, 2019 at 7:46

Tsepo Nkalai

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