Possible Duplicates:
Why does simple C code receive segmentation fault?
Modifying C string constants?
Why does this code generate an access violation?
int main()
{
char* myString = "5";
*myString = 'e'; // Crash
return 0;
}
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0xC0DEFACE
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Before answering that question, let me ask you one: why did you choose to enclose the '5' in double-quotes and the 'e' in single-quotes? – Sev Jun 22 '09 at 07:43
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5This question gets asked about once a week on SO :), give me a minute to find the original and I'll link you to it. – Binary Worrier Jun 22 '09 at 07:43
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1C or C++? Knowing which language you are coding in would be useful. – Daniel Daranas Jun 22 '09 at 07:45
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@Sev: Because one is a string and the other is a character. – sth Jun 22 '09 at 07:45
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1@ Sev: Let me answer that: Because myString is a char *array* (i.e., C-style string), and *myString is the first character in the array. "5" is actually { '5', '\0' }. – DevSolar Jun 22 '09 at 07:46
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@sth: '5' and 'e' both cannot be considered characters? – Sev Jun 22 '09 at 07:46
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5Duplicae of http://stackoverflow.com/questions/164194/why-does-simple-c-code-receive-segmentation-fault – Binary Worrier Jun 22 '09 at 07:46
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"X" is a string ( 'X' + '\0' ) 'X' is just a character – 0xC0DEFACE Jun 22 '09 at 07:47
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@ Daniel Daranas: Does it really make a difference? If you can't tell from the source, usually not. – DevSolar Jun 22 '09 at 07:47
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1Here's one thread: http://stackoverflow.com/questions/1011455/is-it-possible-to-modify-a-string-of-char-in-c/1011481 – Jonathan Maddison Jun 22 '09 at 07:48
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My point was that if the OP initialized it to "e" it would result in the same problem, no? – Sev Jun 22 '09 at 07:48
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@DevSolar: It does make a difference. In one case I can recommend the user to use std::string, in the other, I can recommend an album by The Cure :) – Daniel Daranas Jun 22 '09 at 07:53
3 Answers
5
*mystring is apparently pointing at read-only static memory. C compilers may allocate string literals in read-only storage, which may not be written to at run time.
Avi
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The correct way of writing your code is:
const char* myString = "5";
*myString = 'e'; // Crash
return 0;
You should always consider adding 'const' in such cases, so it's clear that changing this string may cause unspecified behavior.
inazaruk
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