C character array initialization

Question

In the following code:

char *p = "Linux";

Is the memory for "Linux" on the stack or the read only segment of the program?

Refer to question 9 in the article 12 Interesting C Interview Questions and Answers.

Thanks.

@Pubby My instructor for C programming always says its on the stack, hence the doubt. — Jake
– Jake, Commented Nov 27, 2012 at 0:40

David Schwartz · Accepted Answer · 2012-11-27 00:51:05Z

3

The implementation is free to store it wherever it wants. It's a constant, so it can be in read-only memory, but it is not required to be.

My instructor for C programming always says its on the stack, hence the doubt.

He probably means the pointer. Consider:

char *p = "Linux";
p = "Rules";

In the second line, something changed from pointing to "Linux" to pointing to "Rules". That thing that just changed is on the stack.

edited Nov 27, 2012 at 0:51

answered Nov 27, 2012 at 0:38

David Schwartz

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chacham15 · Accepted Answer · 2012-11-27 00:44:07Z

0

As the link says, it is not stored in dynamically allocated memory, rather the memory that is where the code itself lives. I.e. the read only sections. Hence the reason why trying to change it results in a segfault.

answered Nov 27, 2012 at 0:44

chacham15

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