Why does this code fail?

This is a discussion on Why does this code fail? within the C Programming forums, part of the General Programming Boards category; This fails: Code: char *s = "hello"; s[0] = 'H'; but this doesn't: Code: char s[] = "hello"; s[0] = ...

  1. #1
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    Why does this code fail?

    This fails:
    Code:
    char *s = "hello";
    s[0] = 'H';
    but this doesn't:
    Code:
    char s[] = "hello";
    s[0] = 'H';
    or this:
    Code:
    char *s = "hello";
    printf("%c\n", s[0]);
    Besides being a constant pointer, I've always thought char [] was equivalent to char *.

  2. #2
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by Yasir_Malik View Post
    Besides being a constant pointer, I've always thought char [] was equivalent to char *.
    And you have now discovered that it's not.

  3. #3
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    I thought I wrote a reply earlier...

    Can you please describe what you mean by "fails", as I don't see anything that is OBVIOUSLY wrong with your code. But "fails" and "works" are fairly broad terms - "works" could mean that it doesn't crash, but if it comes up with $-1927489231.32 as my "balance" after I've paid this months electrics bill, it may not be quite "working correctly" - so it failed on the next level: operating CORECTLY!

    Also, it would help if you could give a little more code around these code-snippets, because it may be something else that is making it go wrong...

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  4. #4
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by matsp View Post
    Can you please describe what you mean by "fails"
    I assume it crashes. You can't modify a string literal on most platforms.

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    Quote Originally Posted by brewbuck View Post
    I assume it crashes. You can't modify a string literal on most platforms.
    Ah, yes, it crashes on my machine too - for that very reason, the string appears to be in "code" (or is it called "text") memory, so it can't be written to.

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    But why does assigning to an index in char str[] work? Is there something special about char *str that puts the string into the text memory, not into data memory?

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    Code:
    char s[] = "something";
    creates an array of char that is long enough to hold "something", and initialized to that value. Since the variable "s" is in data-memory, and it's containing "something", it can be written to.

    Code:
    char *s = "something"
    creates a pointer s, that points to the constant string "something". Since "something" is a constant, it's not expected to be written to.

    Similarly if you do a function-call:
    Code:
    void blah(char *s)
    {
       ...
    }
    
    ...
        blah("something");
    ...
    blah can't change the input string [and some compilers may catch that by saying "loosing const attribute" or some such].

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    I see now. It goes back to how char s[] is a constant pointer while char *p is just a pointer. Since it can point to anything, it would pointless to have the string "something" floating around; otherwise, there would be a memory leak if I did char *p = "asdf" multiple times in a function.

  9. #9
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by Yasir_Malik View Post
    I see now. It goes back to how char s[] is a constant pointer while char *p is just a pointer. Since it can point to anything, it would pointless to have the string "something" floating around; otherwise, there would be a memory leak if I did char *p = "asdf" multiple times in a function.
    Code:
    char s[] = "hello";
    is the same as
    Code:
    char s[6] = "hello";
    Leaving out the 6 just means that the compiler must infer the size from the thing after the equals.
    I.e. s is a variable that holds an array of 6 chars. Those chars are initialised to the string "hello" (i.e. copied into the buffer), but you're free to overwrite them as you please.
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  10. #10
    Deathray Engineer MacGyver's Avatar
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    Simply put, allocating the array provides the space for the string literal to be automatically copied to. This copy is done automatically behind the scenes. If you use a straight pointer, you end up pointing to wherever the string literal is stored, which is generally in read-only memory.

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    I was thinking about doing this many times:
    Code:
    void func(void)
    {
     char *s = "hello";
    }
    If the compiler allocated memory for s in the data segment. Eventually there would be "hello"s everywhere.
    In addition, s can point to anywhere, so it could point to another char * later on in the code. "hello" would be lost and so would that memory; you can't call delete or free on that. Instead, allocate "hello" once in the text segment.

  12. #12
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by Yasir_Malik View Post
    I see now. It goes back to how char s[] is a constant pointer while char *p is just a pointer.
    char s[] is not a "constant pointer." It is an array. The name of the array acts like a constant pointer in an R-value context. That's the relationship, not more, not less.

    The reason the char s[] version is writable is precisely BECAUSE it is an array.

  13. #13
    Registered User hk_mp5kpdw's Avatar
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    Quote Originally Posted by Yasir_Malik View Post
    I was thinking about doing this many times:
    Code:
    void func(void)
    {
     char *s = "hello";
    }
    If the compiler allocated memory for s in the data segment. Eventually there would be "hello"s everywhere.
    In addition, s can point to anywhere, so it could point to another char * later on in the code. "hello" would be lost and so would that memory; you can't call delete or free on that. Instead, allocate "hello" once in the text segment.
    Memory for s (the actual pointer itself) is allocated from the stack when the function is called. The memory it points to, where the string "hello" resides and the address of which is assigned to s, is in the data segment which is read-only.
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