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Home/ Questions/Q 8107747
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T00:52:39+00:00 2026-06-06T00:52:39+00:00

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

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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?

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  1. Editorial Team
    Editorial Team
    2026-06-06T00:52:40+00:00Added an answer on June 6, 2026 at 12:52 am

    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.

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