7 added 2 characters in body

f s=last[x|x<-0:s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in ss x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now

f s=last[x|x<-0:s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now

f s=last[x|x<-0:s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now

6 added 9 characters in body

f s=last$0:[x|x<s=last[x|x<-0:s,and[x*x<=x*y|y<-s]]  using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y. x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true. we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result. f then works by searching s for an element to satisfy this, and uses 0 as a default. *though he didn't use it for the reasons I'm using it, and he actually got rid of it by now Haskell, 62613938 bytes f s=last$0:[x|x<-s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now

f s=last[x|x<-0:s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now

5 added 90 characters in body

f s=last$0:[x|x<-s,and[x*x<=x*y|y<-s]]  using some comparison magic borrowed from @Zgarb's answeranswer* , namely, x*x<=x*y. x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true. we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result. f then works by searching s for an element to satisfy this, and uses 0 as a default. *though he didn't use it for the reasons I'm using it, and he actually got rid of it by now Haskell, 626139 38 bytes f s=last$0:[x|x<-s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer, namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

f s=last\$0:[x|x<-s,and[x*x<=x*y|y<-s]]


using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now

4 added 7 characters in body
3 added 539 characters in body
2 added 9 characters in body
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