Integer concenation with macro

[nimdays]

Hello,

How to concenate integer correctly with macro ?
I must call it twice here because i can't add something after ","(error)

Code: Select all

#define concat(a,b,c) a##b##c
dim as integer a=10,b=20,c=30,d
d = a concat(*100+,,)b
d = d concat(*100+,,)c
?d  '102030
sleep

Thanks.

[integer]

Hello,

How to concenate integer correctly with macro ?
I must call it twice here because i can't add something after ","(error)

Code: Select all

#define concat(a,b,c) a##b##c
dim as integer a=10,b=20,c=30,d
d = a concat(*100+,,)b
d = d concat(*100+,,)c
?d  '102030
sleep

Thanks.

perhaps this:

Code: Select all

#define concat(a,b,c)  (a*100+b)*100+c
'' OR 
#macro concatenate(a,b,c)
  a*10000 + b*100 + c
#endmacro
''-------------

dim as integer a=10, b=20, c=30, d
d = concat(a,b,c)
print d;"  s/b '102030'"
print
dim as integer x=101,  y=22,  z=3
print concat(x,y,z);" s/b  '1012203'"
print
d = concatenate(a,y,z+1)		''notice the addition inside the macro
print d

[fxm]

To avoid unexpected precedence order of operators, put parentheses systematically inside macro body (even where this is not absolutely necessary) is a good habit.
From the above example:

Code: Select all

#define concat(a,b,c)  (((a)*100+(b))*100+(c))
'' OR
#macro concatenate(a,b,c)
  ( (a)*10000 + (b)*100 + (c) )
#endmacro

[nimdays]

perhaps this:

Code: Select all

#define concat(a,b,c)  (a*100+b)*100+c
'' OR 
#macro concatenate(a,b,c)
  a*10000 + b*100 + c
#endmacro
''-------------

dim as integer a=10, b=20, c=30, d
d = concat(a,b,c)
print d;"  s/b '102030'"
print
dim as integer x=101,  y=22,  z=3
print concat(x,y,z);" s/b  '1012203'"
print
d = concatenate(a,y,z+1)		''notice the addition inside the macro
print d

Thank you both.

To avoid unexpected precedence order of operators, put parentheses systematically inside macro body (even where this is not absolutely necessary) is a good habit.
From the above example:

Code: Select all

#define concat(a,b,c)  (((a)*100+(b))*100+(c))
'' OR
#macro concatenate(a,b,c)
  ( (a)*10000 + (b)*100 + (c) )
#endmacro

[integer]

To avoid unexpected precedence order of operators, put parentheses systematically inside macro body (even where this is not absolutely necessary) is a good habit.
...

Thank you.
That might explain some problems previously experienced with macros.

[fxm]

That might explain some problems previously experienced with macros.

There is already a "WARNING" at description end of #DEFINE page in documentation:

WARNING: When the macro body contains an expression with one operator at least, it may be mandatory to have to surround some terms (parameters, whole body) by parentheses in order to not undergo an unwanted precedence change of operators.

From still the same example, I put in red the mandatory parentheses added because it's a macro and in green the added but optional parentheses:

#define concat(a,b,c) _
( ( ( a ) * 100 + ( b ) ) * 100 + ( c ) )
'' OR
#macro concatenate(a,b,c)
( ( a ) * 10000 + ( b ) * 100 + ( c ) )
#endmacro

[dodicat]

The brackets inside a macro are important.
I had just about given up with this macro to simulate memcpy from crt.bi
As a last resort, before binning everything, I bracketed off the macro.
And then, it seemed to work:

Code: Select all

#include "crt.bi"


#macro memcpyFB(dest,src,size)
scope
   var sz=(size)/sizeof(*(src))
    for n as long=0 to sz-1
        (dest)[n]=(src)[n]
    next
end scope
#endmacro


'udt test
type udt
    as integer i
    as string s
    as long a(45)
end type

dim as udt x(123),y(123),z(123)
x(25).i=13
x(25).s="Hello"
x(25).a(13)=78

memcpyFB(@y(0),@x(0),(ubound(x)-lbound(x)+1)*sizeof(x))

print y(25).i,y(25).s,y(25).a(13),"freebasic"


memcpy(@z(0),@x(0),(ubound(x)-lbound(x)+1)*sizeof(x))

print z(25).i,z(25).s,z(25).a(13),"c runtime"
'===============
'array test
dim as string a(3,8),b(3,8),c(3,8)

dim as long size=((ubound(a,1)-lbound(a,1)+1)*(ubound(a,2)-lbound(a,2)+1)*sizeof(string))

a(3,8)="FreeBASIC"
memcpyFB(@b(0,0),@a(0,0),size)
print b(3,8)

a(3,8)="C runtime"
memcpy(@c(0,0),@a(0,0),size)
print c(3,8)
sleep

 

[fxm]

Indeed, in your above macro example, but without parentheses:

@variable[n]

would be parsed as:

@(variable[n])

due to the operators precedence between '[]' and '@',

but the wanted expression is:

(@variable)[n]