Question

First hallo to everyone, first post here.

Today I found myself wondering if this code is correct, it was written time ago by a friend for a microcontroller and I have no chance to ask him anymore.

The program works correctly, but I can’t really decide if it works by luck or not (don’t belive in luck with programming).
I don’t even know if the title of this post is correct, sorry if it could be misleading.
The code:

char  *textStatusErrorMessage( unsigned int codeStatus )
{
    switch ( codeStatus ) {

    case STATUS_1:
        return ( (char  *) " Status: 1 " );
        break;     
    case STATUS_2:
        return ( (char  *) " Status: 2 " );
        break;      
    default:
        sprintf( tmpBuf, " UNKNOWN STATUS   %03d ", codeStatus );
        return ( (char  *) tmpBuf ); //tmpBuf is global
        break;
    }
}

more precisely this syntax is kind of obscure to me.

return ( (char  *) " Status: 1 " );

It returns a char * of what ? where is the “Status: 1 ” string saved ? heap/stack???

As it’s implemented the string is in the scope of the function and I’m assuming that after leaving the function with the return statement I have no control of what is gonna be written in the pointer returned by the function itself.

From the way I see it I would have had a global array with the different possible string-options and returned a pointer to the correct one selected by the CASE. So I know the pointer I return is to a well defined memory area.

So is this code wrong or not? Which is the most correct solution?

Thanks
DAN

Answer 1

In theory, this code is valid; string literals (e.g. " Status: 1 ") have a type of char[], and a lifetime equal to that of the whole program.

However, there are a few issues:

1. In the string literal cases, if anything tries to modify the returned string, then undefined behaviour occurs. Really, this function should return a const char * instead.

2. It’s not clear where tmpBuf is declared, nor whether it points to sufficient memory to hold the data being written to it.

3. There is only one tmpBuf, so every time this function is called, tmpBuf will be overwritten. This is likely to cause hard-to-track bugs.

4. The casts are unnecessary.