Modifying a C string: access violation [duplicate]

Question

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;
}

This 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, Commented Jun 22, 2009 at 7:43
C or C++? Knowing which language you are coding in would be useful. — Daniel Daranas, Commented Jun 22, 2009 at 7:45
@ 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, Commented Jun 22, 2009 at 7:46

Avi · Accepted Answer · 2009-06-22 07:44:23Z

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.

answered Jun 22, 2009 at 7:44

Avi

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DevSolar · Accepted Answer · 2009-06-22 07:44:37Z

2

String literals are considered constant.

answered Jun 22, 2009 at 7:44

DevSolar

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inazaruk · Accepted Answer · 2009-06-22 07:51:10Z

0

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.

answered Jun 22, 2009 at 7:51

inazaruk

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