Array of pointers, changing pointer elements

Question

If I declare an array of pointers like that:

    char* arr[5] = {"Mercury",
    "Mercury",
    "Venus",
    "Earth",
    "EArth"};

Can I then change sings in those pointers? I have tried doing something like that

*(*(arr + 1) + 1) = 'i';

but it doesn't work, I get memory dump. Is there a way to do that or I have to declare it differently?

Things such as "Mercury" are string literals. Trying to modify a string literal results in undefined behaviour, most of the time a seg fault. — Jabberwocky
– Jabberwocky, Commented Apr 28, 2020 at 13:41
Each of the 5 pointers in the array points to a string literal. These are constant data, and cannot be changed. — Adrian Mole
– Adrian Mole, Commented Apr 28, 2020 at 13:41
String literals in C are usually stored in read-only memory. If you allocate read-write memory, either on the stack by declaring an array, or on the heap with malloc, you will be able to modify the strings. — Program man
– Program man, Commented Apr 28, 2020 at 13:41
OT: *(*(arr + 1) + 1) = 'i';--> arr[1][1] = 'i'; but it is still illegal code — 4386427
– 4386427, Commented Apr 28, 2020 at 13:55
You CAN swap pointers, e.g. char *tmp = arr[2]; arr[2] = arr[0]; arr[0] = tmp; would be fine. That would result in the ordering "Venus", "Mars", "Mercury", "Earth", "Pluto" — David C. Rankin
– David C. Rankin, Commented Apr 28, 2020 at 14:04

RobertS supports Monica Cellio · Accepted Answer · 2020-04-28 14:03:30Z

1

Is there a way to do that or I have to declare it differently?

char* arr[5] = {"Mercury", "Mars", "Venus", "Earth", "Pluto"};

arr is an array of 5 char pointers to string literals. Any attempt to modify a string literal invokes undefined behavior, so you can´t modify them.

If you want to modify the content you need f.e. to define them as two-dimensional array of chars:

char arr[5][10] = {{"Mercury"}, {"Mars"}, {"Venus"}, {"Earth"}, {"Pluto"}};

and use

strcpy(arr[0], "Uranus");

size_t len = strlen(arr[0]);  
for(size_t i = 9; i > (len + 1); i--)  // To remove all left characters from prev. string.
{
    a[0][i] = '\0';
}

edited Apr 28, 2020 at 14:03

answered Apr 28, 2020 at 13:46

RobertS supports Monica Cellio

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