The software is distributed under CC-GNU GPL license.

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The software is distributed under CC-GNU GPL license.

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The software is distributed under CC-GNU GPL license.

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• MPI,
• OpenMPI,
• pthreads,
• parallel Sudoku solver and generator.

The software is distributed under CC-GNU GPL license.

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In part 1 we introduced the const keyword. Today we’ll talk about constant pointers.

Let’s say you want to define a constant pointer. Which of the following declarations will you use?

The answer is: the second or the third. The first declaration defines a pointer to a constant integer and the third one defines a constant pointer to a constant integer.

Another interesting case is when defining a constant char array (credits go to Ulrich Drepper, link via RazvanD):

Although this code will give a warning (passing `const char *’ as argument 1 of `strcpy(char *, const char *)’ discards qualifiers is the exact message on Dev-C++), it will run, because s is allocated in the heap, so it is treated much like a pointer. You can force the value to be constant by adding the static keyword, wich will force the compiler to allocate s in read-only memory:

First of all, let’s see what the const modifier means in the C standard. Basically, a const variable is one who’s value can’t be changed. Actually, things are not so simple – as we’ll see later, you can change a constant variable. The standard states that “If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined.” On some architectures, constant variables are put in a special section (sometimes called .rodata – from Read Only DATA) of the program by the compiler.

If you want to define a constant in C/C++, you can write:

Both declarations mean the same thing: define a new integer with a fixed value. So, if you want to change the value of v1, how would you do it? By using pointers:

Of course, you shouldn’t do that, as the are no guarantees that the result will be what you expect it to be, but with most compilers, v1 will be 5 after running the code presented above.

You can use the same code in userspace programs by replacing printk with printf. And just in case you’re wondering what’s with the empty do-while, you might want to take a look at this older article.

Fun, huh? Let’s try one more time:

a pointer to an array of functions
|  receiving a pointer to a function
|               |  receiving an int
|               |  returning a pointer to a function
|               |               |  receiving an array of pointer to int
|               |               |  returning int
|            an int
|  returning a pointer to a function
|               |  receiving a pointer to a function
|               |               |  receiving a pointer to an array of int
|               |               |            an int
|               |               |  returning a pointer to int
|               |  returning a function
|               |               |  receiving int
|               |               |  returning int

It does sounded quite tricky to me. Still, he managed to write it down in just 10 minutes. The secret? Cdecl. Cdecl reads files for statements which should be translated to C or C++. Six different statement are supported: declare composes a C type-declaration, cast composes a C type-cast, and so on. The grammar is described in full in the man page.

And here is the result for the function described above:

Well, in C it’s pretty simple. All you have to do is to declare an union, like this:

This tells the compiler that both f and bits should be hold in the same memory zone. This is pretty useful to save some memory, if you know for sure that you wown’t need f and bits in the same time.

As a side effect, if you write a value in f then read bits, you will have all the bits represented in the IEEE 754 standard. Now you can use bit operations (& , | , ^) with your number, you can extract the different parts of the number, etc. Do keep in mind that this is bad programming practice.

This basicly means that the code is executed exactly once, so the first idea is that do{}while(0); is useless.
In fact, there are a number of reasons for writing the macros this way:

• Empty statements give warnings from the compilers, so rather than writing
  #define FOO

  you might want to write

  #define FOO do{}while(0)
• It gives you a basic block in which to declare local variables. You could simply use curly brackets, but this could cause serious problems in conditional statements. Let’s take the following example:
  #define exch(a,b) {int t; t = a; a = b; b = t;}

  You then use the macro in the following code:

  if(a[i]<a [i+1])
      exch(a[i],a[i+1]);
  else
      ready = 1;

  This is what the copiler will get:

  if(a[i]<a [i+1]){
      int t;
      t = a[i];
      a[i] = a[i+1];
      a[i+1] = t;
  };
  else
      ready = 1;

  This is interpreted as an if statement whithout else and you will get an error like “else without matching if”. However, if you write the macro like this:

  #define exch(a,b) do{int t; t = a; a = b; b = t;}while(0)

  the code will behave as expected.

You can find more informations about this in the Kernelnewbies FAQ.