Create a random number and check if it is odd or even

Question

I am looking to create a batch file that, when ran, creates a random integer from 0 to 100 and then runs it through an if statement that checks whether it is odd or even. I\'ve

Answer 1

The following prints a random number between 0 and 99, and prints whether it's even or odd:

@echo off

set /a num = %random% %% 100
echo %num%
set /a odd = num %% 2
if %odd% EQU 1 (echo odd) else (echo even)

Answer 2

@ECHO OFF
SETLOCAL
set /a num=%random% %%100 +1
SET /a nummod2=num %% 2
IF %nummod2% == 0 (ECHO %num% is even) ELSE (ECHO %num% is odd)

GOTO :EOF

Conventional IF syntax is if [not] operand1==operand2 somethingtodo where operand1 and operand2 are both strings. If the strings contain separators like Space,Tab or other characters that have a special meaning to batch then the string must be "enclosed in quotes".

Fundamentally, if compares strings. The operator must be one of a fixed set of operators [== equ neq gtr geq lss leq] hence cmd was objecting to num where it expected an operator.

A calculation cannot be performed within a if statement's parameters.

%% is required to perform the mod operation, since % is a special character that itself needs to be escaped by a %.

Note that { and } are ordinary characters with no special meaning to batch and remarks must follow

rem remark string

There is a commonly-used exploit which is

::comment

actually, a broken label, which can have unforeseen side-effects (like ending a code-block)