Segmentation fault when trying to assign new value to my pointer

Question

char *c = "H'eLo";
for (int i = 0; i < 5; i++)
{
    if (isupper(c[i]))
    {
        c[i] = tolower(c[i]);
        printf("%c \n", c[i]);
    }
}

So I am running this code, and trying to figure out what I am doing wrong. What is the reason for the segmentation fault?

Change `char *c = ...` to `char c[] = ...`. The reason the pointer version doesn't work is you have a pointer to a string constant, which cannot be modified. Changing it to a statically initialized array will fix it. — Tom Karzes, Jun 09 '20 at 16:22
You make `c` point to a string constant. Then you try to modify the thing `c` points to. But, by definition, you can't modify a constant. — David Schwartz, Jun 09 '20 at 22:31

Mapromu · Accepted Answer · 2020-06-09T22:57:44.730

-1

If you want to modify the string, you should allocate it dinamically just like this

    char *c = malloc(5*sizeof(char));

Then, pass the string with strcpy

    strcpy(c, "H'eLo");

Or you also can do it with the function strdup, which is the easiest option

    char *c = strdup("H'eLo");

These examples should work, because C doesn't let you modify strings literals

edited Jun 09 '20 at 22:57

answered Jun 09 '20 at 16:20

Mapromu

It simply isn't allowed to modify string literals in C, so it's not a matter of recommendation, but instead requirement :-) Even if it doesn't crash sometimes, it gives the program undefined behaviour, so it can fail in baffling ways later. – underscore_d Jun 09 '20 at 16:22
Thanks !! I will add this to the answer – Mapromu Jun 09 '20 at 16:24
2

You have unbalanced parentheses. – Tom Karzes Jun 09 '20 at 16:41
char *c = strdup("Hello World") – pm100 Jun 09 '20 at 22:24
This writes out of bounds. as well as using redundant cast – M.M Jun 09 '20 at 22:26
Thanks for the comments guys !! – Mapromu Jun 09 '20 at 22:55

1 Answers1