Typecasting string and strdup

Question

If an input const string is being modified in some way (which is resulting in C compiler warning), what is the best way to handle it - typecasting it to a new variable and then using it OR duplicating it and using it and then freeing it. Or is there any other way to handle this type of scenario. please suggest. Any help would be appreciated.

//Typecasting

const char * s1;
char * s2 = (char *)s1;

//Duplicate and free

const char * s1;
char * s2  = strdup( s1 );
free(s2)

EDIT: It is a C compiler; not C++. I am not sure whether in typecasting, s2 will be a new copy of string s1 or will it be pointing to the original string s1?

Thanks for the answers. I have one more doubt-

const char * c1;
const char * c2 = c1;

Is the above assignment valid?

It depends on the details of the reason you need to modify a const string. — David Schwartz, Oct 11 '11 at 06:35

David Heffernan · Accepted Answer · 2011-10-11T07:25:19.057

Casting away const here might shut the compiler up but will lead to runtime failures. Make a copy of the string and work on that.

Casting away const does not copy the contents of the memory. It just creates a pointer to the same memory and tells the compiler that it can go right ahead and write to that memory. If the memory is read only you have a protection fault. More seriously you can have correctness problems which are hard to debug. Don't cast away const.

Of course, if you need to modify a variable and have those modifications visible to the caller, then you should not make it const in the first place. On the other hand, if the modifications are meant to be private to the function, then duplication of the const parameter is best.

As a broad rule you should attempt to avoid casts if at all possible. Casts are one of the most frequent sources of errors.

There's not much more to add. For example string literals reside in read only memory and casting away const will result in protection faults when you try to write to that memory. Don't cast away const. — David Heffernan, Oct 11 '11 at 06:59

Mike Seymour · Answer 2 · 2011-10-11T07:13:47.977

2

If the array that s1 points to is const then you must not modify it; doing so gives undefined behaviour. In that case, if you need to modify it, then you will have to take a copy. UPDATE: Typecasting will give you a pointer to the same array, which is still const, so you still must not modify (even though trying to do so no longer gives a compile error).

If you can guarantee that the array is not const, then you could modify it by casting away the const from the pointer. In that case, it would be less error-prone if you can enforce that guarantee by not adding a const qualifier to s1 in the first place.

edited Oct 11 '11 at 07:13

answered Oct 11 '11 at 06:57

Mike Seymour

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s1 is definitely const in my case. So are you saying that strdup is a better option than typecasting? – user32262 Oct 11 '11 at 07:03
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That's exactly what we are all telling you! – David Heffernan Oct 11 '11 at 07:05
@user32262: yes, typecasting the pointer will give you a pointer to the same array, which you still mustn't modify. I've updated my answer to clarify. – Mike Seymour Oct 11 '11 at 07:14

score 0 · Answer 3 · answered Oct 11 '11 at 06:34

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If your input is a const char* you definitely should listen to the compiler warning and leave it alone.

answered Oct 11 '11 at 06:34

Ayjay

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Typecasting string and strdup

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