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Q&A

How to write a macro that discards the const qualifier, for any type?

+1
−0

How to write a macro that discards the const qualifier, for any type?

I hope some combination of typeof and a cast will do, but haven't found the combination.

I tried this, without luck:

#define discard_const(_x)  ((typeof(_x + 0)) (_x))

struct t {
    char *s;
};

char *
foo(const struct t *r)
{
    return discard_const(r->s);
}

I also tried: #define discard_const(_x) ({__auto_type _y = (_x); _y;}), but not luck

The objective is that if I cast to (char *), I fear that if the type of s changes in the future, the cast might silence any warnings, so I want the type to be automatically calculated, and I only want to discard const.

Of course, with pragmas I may disable any const warnings for that specific line, but a macro that does it within the language would be nicer.

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2 comment threads

Most of the time, you don't (3 comments)
The variable you're applying `discard_const` to is not const. (4 comments)

1 answer

+4
−0

Ignoring the numerous forms of undefined behavior that casting away const might invoke, the blunt but simple and standard solution is just to cast to (void*).

char* foo (const char* str)
{
    return (void*)str;
}

This is far more portable than gcc extensions like typeof.

(C23 might introduce various type-related features similar to _Generic so there might be more elegant ways coming soon.)

The only application for these kind of dirty casts is pretty much when dealing with broken API functions that take const-correct parameters but return a non-const pointer, like for example:

char *strstr (const char *s1, const char *s2);

This function is broken by design - always was. Correct design would be to return an index instead of a pointer. But if you are stuck with a broken API such as this one and have to implement it, you have to take shortcuts like casting away const.

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1 comment thread

Type generic programming. (1 comment)

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