Question

In c++

lets say i have a variable char lets call it

char variable;

and i want to test it if it is any of these

'a', 'b', or 'c'

after the user inputs a char
will the below snip of code work as a shorcut

if(variable== 'a'|| 'b' || 'c') cout << "your variable is one of those";

or do i have to do it the long way like below

if(variable== 'a'|| variable =='b' || variable== 'c') cout << "your variable is one of those";

Answer 1

Or lets say i made this function that takes char c and returns true if it is a number 0-9.
Do I absolutely have to write it like that? all of it on the if statement?


bool boolisnum( char c)
{
if (c=='0'||c=='1'||c=='2'||c=='3'||c=='4'||c=='5'||c=='6'||c=='7'||c=='8'||c=='9') return true;
else return false;

};


can

if (c=='0'||c=='1'||c=='2'||c=='3'||c=='4'||c=='5'||c=='6'||c=='7'||c=='8'||c=='9')

be reduced to

if (c=='0'|| '1'||'2'||'3'||'4'||'5'||'6'||'7'||'8'||'9')

Answer 2

Use switch case.

Answer 3

char c = 'a';
if (c=='0'|| '1'||'2'||'3'||'4'||'5'||'6'||'7'||'8'||'9')
cout << "test";
that code segment still outputs "test" no matter what value c is. I think the reasoning behind this is that you can think of characters as integers.

if(CONSTANT) executes the conditional as long as the number doesn't evaluate to 0 (false). if('6') is the same thing as saying if(54). Look at the ascii table: http://www.asciitable.com/

bool b = true;
if(1) and if(54) and if(-1) and if(b) all execute the conditional. Use a switch or the conditional operator. Either way, write it out.

Answer 4

I know for sure works.

bool boolisnum( char c)
{
if (c=='0'||c=='1'||c=='2'||c=='3'||c=='4'||c=='5'||c=='6'||c=='7'||c=='8'||c=='9') return true;
else return false;

};

if i where to type in main
boolisnum('4');

it will evaluate to true Note that idk what

boolisnum(4);

would do. It probably would not compile since 4 is of type int not char.

I am just looking to make my code shorter by not putting all those c==' 's.

oh and if
boolisnum('b');
it evaluates as false.

Answer 5

#include <cctype>
isdigit()
http://www.cplusplus.com/reference/cctype/isdigit/

Answer 6

or, switch(c){
case '1':
case '2':
...
case '0':
etc.
break;
}

Answer 7

switch makes it longer. The whole point is to make my code shorter.

Oh and os is messing up.

Answer 8

I literally just gave you a single function. It doesn't get shorter than that

Answer 9

bool boolisnum( char c){
if (isdigit(c))
return true;
return false;
}

Answer 10

yes thank you that does work now i wanna know

is

if (c=='0'|| '1'||'2'||'3'||'4'||'5'||'6'||'7'||'8'||'9')

same as

if (c=='0'||c=='1'||c=='2'||c=='3'||c=='4'||c=='5'||c=='6'||c=='7'||c=='8'||c=='9')

Answer 11

I explained it in my first post. It isn't the same thing

Answer 12

It'll always evaluate to true no matter what

Answer 13

darn so i do gotta use

c=='0'||c=='1'||c=='2'||c=='3'||c=='4'||c=='5'||c=='6'||c=='7'||c=='8'||c=='9'

Answer 14

You don't gotta use anything when there's a function written for you.

If you refuse to use perfectly find functions, yeah, you have to do that...

Answer 15

I am just saying as an example. I know there is one that checks for if its upper of lower case.

BUt There wont be one for every single one of my needs :).

Answer 16

lol the STL is pretty comprehensive

check http://www.cplusplus.com/reference/cctype/ for more. Read about how chars/ascii work in relation to integers.

remember if(some constant that isn't 0) always evaluates to true

Answer 17

it is a huge library. And sometimes it be quicker making my function instead of looking for it :P

Answer 18

and it's a great learning experience :D

Answer 19

to reiterate why
if (c=='0'|| '1'||'2'||'3'||'4'||'5'||'6'||'7'||'8'||'9')
will always evaluate to "true", is because c++ interprets zero as false, and all non-zero values as true. No char has an ascii value of zero, as binary zero in ascii is the NUL value. For instance

if (false || 1)
will always evaluate to "true", since the second statement, 1, counts as true. equally it could be any non-zero number.

if (false || -534.7)
will also evaluate to "true"

As far as the compiler is concerned, this is the same as
if (false || true)

only something like

if (false || 0)
will evaluate to "false", since zero counts as false. Since the char '0' has an ascii value of 48 in decimal, the statement

if (false || '0')
is equivalent to
if (false || 48)
which, as I explained above, is equivalent to
if (false || true)
which will, of course, evaluate to "true"

Answer 20

'a'=='d' will evaluate as false
'a'=='a' will evaluate as true i know that much.

Answer 21

there's no equivalence being done here. The characters are getting interpreted by their INTEGER ascii values.

if('a') is the same thing as if(65) or whatever

Answer 22

idk what it does but those statements will work i have compiled them and it does work.

Answer 23

use full sentences because I don't understand you.

if(c=='0'||'1'...)
will not work...

Answer 24

Okay, ASCII was ordered the way it was for a reason.
```
# is digit
if '0' <= variable <= '9':

# is a,b, or c
if 'a' <= variable <= 'c':
```
Look at the chart.

Answer 25

I suppose since this is C++ you should be using
`if (` ... `) {` ... `}` instead, but the point still stands.

Answer 26

Why reinvent the wheel? Regex is a great way to check input against a pattern in any language. http://www.cplusplus.com/reference/regex/regex_search/