const char* and free()

问题:

Given the next code example, I'm unable to free the parameter const char* expression:

// removes whitespace from a characterarray
char* removewhitespace(const char* expression, int length)
{
    int i = 0, j = 0;
    char* filtered;
    filtered = (char*)malloc(sizeof(char) * length);
    while(*(expression + i) != '\0')
    {
        if(!(*(expression + i) == ' '))
        {
            *(filtered + j) = *(expression + i);
            j++;
        }
        i++;
    }
    filtered[j] = '\0';
    free(expression); //this doesn't seem to work
    return filtered;
}

Before I return this function, I try to free the data in the expression parameter but I can't seem to free it.
I think it is probably because it is a constant, but I learned that a character array in C always should be a constant.

The error message I get is at the line with free(expression) and the message is:
expected void* but argument is of type const char * - compiler error

How do I discard the memory that the data expression contains?

回答1:

If it's a constant, that implies that it shouldn't be deleted, because it wasn't allocated: the caller could easily have passed a stack-based array to your method, and freeing it would be a bad thing.

The most common convention is that whatever code allocates some data should free the data. It's really not your routine's responsibility to free that argument since it has no way of knowing if that's even a legitimate thing to do.

回答2:

If you really want to have a function like this - taking a heap pointer, freeing it and in spite returning another heap pointer - you should document it very well.

Besides, you shouldn't make expression a const char*, because that doesn't match, as you see. char* should be fine, however.

回答3:

going on directly straight forward with the question is free((void*)expression); that should be the enough to resolved the compiling error. but, as already others have state you should let who is calling your function to manage the memory