Segmentation fault in c array, access to individual char in a string

Question

I am attempting to manipulate individual characters in a string, in this case change 4th 'a' to a 'b'.

string password = "aaaaa";
printf("password: %s\n",password);

int j = 'b';
password[3] = (char) j;
printf("password: %s\n",password);

this returns:

password: aaaaa

Segmentation fault

One last note: in the first line I declare 'string' like a variable. This contrivance is allowed by the CS50 library - It should work and I have used it in the past.

Thank you in advance.

Thank you for your comments - I now updated the tag. Yes I am doing the excellent cs50 and have included #include — Tikhon, Dec 16 '18 at 05:16

score 4 · Accepted Answer · answered Dec 16 '18 at 05:10

4

"aaaaa"; is a String Literal which is immutable on most systems, so password[3] = (char) j; attempts of modify an immutable object resulting in a SegFault.

Instead,

char password[] = "aaaaa";

Presuming your "string" is a typedef of char* using a compound literal allows the same result, e.g.:

string password = (char[]){"aaaaa"};

answered Dec 16 '18 at 05:10

David C. Rankin

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That's done it, thank you very much. – Tikhon Dec 16 '18 at 05:13
Sure, glad to help. Good luck with your coding. – David C. Rankin Dec 16 '18 at 05:13

Segmentation fault in c array, access to individual char in a string

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