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Old 01-31-2005, 10:41 PM   #1
Bluesuperman
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C++ question #define ...


Hello,

I am trying to learn C++ programming, I read through some online tut... and picked up a book called "How To Program C++" by Deitel and Deitel.

But I must of missed something where ... because I am not sure I understand the following statements.

--snip--
#define ARRAY_SIZE(a) (sizeof(a) / sizeof(a[0]))
#define CTRLD 4

char *choices[] = {
"Choice 1",
"Choice 2",
"Choice 3",
"Choice 4",
"Exit",
};

n_choices = ARRAY_SIZE(choices);
--snip--


So this means anywhere I use "CTRLD" in the program it will be replaced by a init 4 ?

Next is where it gets confused, using define creates constants .. which can not be modified correct ?

So ARRAY_SIZE(a) points to (sizeof(a) / sizeof(a[0])) ... so it is like a place holder ? because in the program ARRAY_SIZE gets passed an array and simple returns the size of the array divided by 1 which is the length of the data is a[0] ?

AHHHH ......

Michael.
 
Old 02-01-2005, 12:34 AM   #2
Dark_Helmet
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Quote:
#define CTRLD 4
...
So this means anywhere I use "CTRLD" in the program it will be replaced by a init 4 ?
Pretty much. It means that whenever CTRLD is encountered in the code, it is replaced by a literal 4. How that "4" is interpreted entirely depends on the context it's used in. In almost all cases, yes, it will be considered an integer.

Quote:
#define ARRAY_SIZE(a) (sizeof(a) / sizeof(a[0]))
...
So ARRAY_SIZE(a) points to (sizeof(a) / sizeof(a[0])) ... so it is like a place holder ?
Not quite sure what you mean by a place holder. When a #define uses parentheses as in this case, you can think of it as defining a mini-function. The "a" in the #define is replaced with the given value in the code. So, for example, using the code you provided
Code:
ARRAY_SIZE(choices)
is equivalent to
Code:
(sizeof(choices) / sizeof(choices[0]))
Again, that's placed directly into the text of the source during the preprocessor phase. The value of the expression is the number of elements in the array, because your dividing the size of the entire array (sizeof(choices)) by the size of each element (sizeof(choices[0])). In this case, that evaluates to 5.

The choices array is made up of character pointers (each of which consumes 4 bytes of data). The array's size as a whole is the sum of all its elements. "Choice 1" is 4 bytes, "Choice 2" is 4 bytes, etc., for a total of 20 bytes for the entire array.
 
Old 02-01-2005, 06:55 AM   #3
pvs
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try #define ARRAY_SIZE(a) (sizeof((a)) / sizeof((a)[0]))
 
Old 02-01-2005, 08:27 AM   #4
bigearsbilly
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What happens is, the c file goes through the macro processor (man m4 i believe)
before compilation and all defines are replaced by the said text.

It's not rocket science, it's very basic and simple!

If you use the -E option it will just show you the pre-processor steps.


observe:
Note: this is not a valid C file, but that's OK because this happens
before actual compilation.

Code:
billym.primadtpdev>cat 1.c
#define ARRAY_SIZE(a) (sizeof(a) / sizeof(a[0]))
#define CTRLD 4

char *choices[] = {
    "Choice 1",
    "Choice 2",
    "Choice 3",
    "Choice 4",
    "Exit",
};

ctrld = CTRLD

n_choices = ARRAY_SIZE(choices);
through gcc -E
Code:
billym.primadtpdev>gcc -E 1.c
# 1 "1.c"



char *choices[] = {
    "Choice 1",
    "Choice 2",
    "Choice 3",
    "Choice 4",
    "Exit",
};

ctrld = 4 

n_choices = (sizeof( choices ) / sizeof( choices [0])) ;
good eh?
 
Old 02-01-2005, 09:49 AM   #5
michael_util
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Thanks for all the replies ...

Michael.
 
  


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