EXPLAIN THIS CODE...

[Former Member]

DATA: result TYPE i,

n1 TYPE i VALUE 5,

n2 TYPE i VALUE 6.

DEFINE operation.

result = &1 &2 &3.

output &1 &2 &3 result.

END-OF-DEFINITION.

DEFINE output.

write: / 'The result of &1 &2 &3 is', &4.

END-OF-DEFINITION.

operation 4 + 3.

operation 2 ** 7.

operation n2 - n1.

[Former Member]

Hi,

DATA: result TYPE i,

n1 TYPE i VALUE 5,

n2 TYPE i VALUE 6.

• <b>This is macro definition.</b>

DEFINE operation.

result = &1 &2 &3. <b>"&1 &2 &3 are place holders</b>

output &1 &2 &3 result. " <b>Calling the macro output</b>

END-OF-DEFINITION.

• <b>This is another macro (output) definition</b>

DEFINE output.

write: / 'The result of &1 &2 &3 is', &4. " <b>4 place holders passed to this macro</b>

END-OF-DEFINITION.

<b>*Clling the macro operation 3 times with three different *operation( addition, power, subtraction )</b>operation 4 + 3.

operation 2 ** 7.

operation n2 - n1.

Regards,

[Former Member]

HI

[Former Member]

Hi srinivas..

that is a macro for which u r passing values

<b>for the first call Operation 4 + 3.</b>

Operation MAcro..

&1 = 4

&2 = +

&3 = 3.

output Macro..

&4 is result....

<b>for the second call Operation 2 ** 3.</b>

&1 = 2

&2 = **

&3 = 3.

output Macro..

&4 is result..

[Former Member]

Hi,

u hav declared 3 variables n1 n2 result. and there are 2 macros defined in the code operation and output.

operation 4 + 3.

the above statement inserts the values in the place holders and after calculating then it places the values in macros result.

regards,

kiran kumar k

[Santosh_Kezkhepatmelathil]

hi,

you have created two Macros operation and output.

and then when you execute the below code their respective macros will get executed

operation 4 + 3.

operation 2 ** 7.

operation n2 - n1.

for example : operation 4 + 3. will execute operator macro in which it Contains the output macro and the result will be displayed.

regards,.

santosh.