Segmentation fault when changing a string

Code:

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#include<stdio.h>
int main(){
  char *p = "Abc";
  *p = 'a';
  printf("%s\n", p);
  return 0;
}

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It received SIGSEGV on the 3rd line. But if I use

Code:

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char p[] = "Abc";

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, it goes well.
So why I can't change *p when p is a pointer to a string? And, it's all right if I change it using gdb when I debug it.

Quote:

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Originally Posted by lilydjwg

So why I can't change *p when p is a pointer to a string?

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Because it is not allowed to change the contents of a string literal, and attempts to do so result in undefined behaviour.

Quote:

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Originally Posted by laserlight

Because it is not allowed to change the contents of a string literal, and attempts to do so result in undefined behaviour.

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If so, why can I when in gdb or use char p[] instead of char *p? What's the difference between them?

Code:

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char *p = "abcde";
char p[] = "abcde";

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In the first statement the address of the first element of the string literal "abcde" is assigned to p. This string literal is constant.

The second case is an array assignment. Arrays can be changed. The second assignment is equal to:

Code:

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char p[] = {'a', 'b', 'c', 'd', 'e', '\0'};

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Code:

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char p

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is one char - not array

Quote:

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Originally Posted by vart

Code:

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char p

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is one char - not array

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That's true: it should be:

Code:

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char p[] = {'a', 'b', 'c', 'd', 'e', '\0'};

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Sorry for the inconvenience, but I adapted the original post.

hilarius, so you mean two following two lines are the same, right?

Code:

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char *p = "abc";
const char *p = "abc";

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And, if you know C++, is it the reason that g++ will warn char *p = "abc" for p is not declared as const?