In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
e.g:
for a = 0, b = 3 and c = 2:
a < b < c = 1
(a < b) && (b < c) = 0
when typed in C.
72  4302 
"Paminu" <ja******@asd.c om> wrote in message
news:di******** **@news.net.uni-c.dk... In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
e.g:
for a = 0, b = 3 and c = 2:
a < b < c = 1
(a < b) && (b < c) = 0
when typed in C.
(0 < 0) && (0 < 0)
false && false = false according to the truth table for logical AND
"pemo" <us************ @gmail.com> wrote in message
news:di******** **@news.ox.ac.u k...
"Paminu" <ja******@asd.c om> wrote in message
news:di******** **@news.net.uni-c.dk... In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
e.g:
for a = 0, b = 3 and c = 2:
a < b < c = 1
(a < b) && (b < c) = 0
when typed in C.
(0 < 0) && (0 < 0)
false && false = false according to the truth table for logical AND
Oops - sorry - bit too quick there!
(0 < 3) && (3 < 2)
true && false = false - logical AND requires both operands to be true to get
a true result
pemo wrote:
"pemo" <us************ @gmail.com> wrote in message
news:di******** **@news.ox.ac.u k...
"Paminu" <ja******@asd.c om> wrote in message
news:di******** **@news.net.uni-c.dk... In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
e.g:
for a = 0, b = 3 and c = 2:
a < b < c = 1
(a < b) && (b < c) = 0
when typed in C.
(0 < 0) && (0 < 0)
false && false = false according to the truth table for logical AND
Oops - sorry - bit too quick there!
(0 < 3) && (3 < 2)
true && false = false - logical AND requires both operands to be true to
get a true result
Yes but then what about
0 < 3 < 2
how is that evaluated?
"Paminu" <ja******@asd.c om> wrote in message
news:di******** **@news.net.uni-c.dk... pemo wrote:
<snip>
Yes but then what about
0 < 3 < 2
how is that evaluated?
Associativity - what happens when operators have the same precedence.
< = left to right
So, you've got ((0 < 3) < 2)
which resolves to (1 < 2)
1 [true] (as 0 IS less than 3)
which resolves to 1 (true)
I think the problem is the order is from left to right,
the value of a<b is TRUE, which equal to 1, then 1 < c is also TRUE, so
the value of a<b<c is TRUE;
but in the (a<b)&&(b<c)
a<b is TRUE ,1
b<c is FALSE, 0
so they equal to 1&&0 ,means 0
Paminu wrote: In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
e.g:
for a = 0, b = 3 and c = 2:
a < b < c = 1
(a < b) && (b < c) = 0
when typed in C.
Remember that C has no boolean type. The (binary) "<" operator returns 0
(false) or 1 (true), so `0 < 3' is 1 and `1 < 2' is 1.
August
Paminu wrote:
In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
Because
(a < b < c)
means
(((a < b) != 0) < c)
--
pete
pete wrote: Paminu wrote:
In math this expression:
(a < b) && (b < c)
would be described as:
a < b < c
But why is it that in C these two expressions evaluate to something
different for the same values of a, b and c?
Because
(a < b < c)
means
(((a < b) != 0) < c)
....or
((((a < b) != 0) < c) != 0)
or why not
(((((a < b) != 0) < c) != 0) != 0)
and so on... ;-)
August
August Karlstrom <fu********@com hem.se> writes:
[...] Remember that C has no boolean type. The (binary) "<" operator returns
0 (false) or 1 (true), so `0 < 3' is 1 and `1 < 2' is 1.
C has had a boolean type for 6 years; C99 added _Bool. It's true that
support for C99 is not yet universal, though.
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
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