What is the operator <? = In C ++

Possible duplicate:
What does the> operator mean? =?

I was wondering what the operator <?=is that I see in C ++ code. It does not compile on my system, but I am sure that it is correct. It is used as follows:a <?= something;

Also, what do I need to do to compile it?

EDIT POST:

Please check out this source code (this is a solution to the Square Fields problem (Problem B) winner of the Google Code Cork Competition practice - see here ). In some other representations, I saw the same character model.

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<numeric>
#include<cmath> using namespace std;

#define ALL(t) t.begin(),t.end()
#define FOR(i,n) for (int i=0; i<(int)(n); i++)
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) typedef vector<int> vi; typedef long long int64;

int t[15][1<<15]; int main() {   int N;cin>>N;   for(int c=1;c<=N;c++){
    int n,k,x[30],y[30];
    cin>>n>>k;
    FOR(i,n)cin>>x[i]>>y[i];
    FOR(take,1<<n)if(take){
      int minx=1000000,maxx=-1,miny=1000000,maxy=-1;
      FOR(i,n)if(take&1<<i)minx<?=x[i],maxx>?=x[i],miny<?=y[i],maxy>?=y[i];
      t[1][take]=(maxx-minx)>?(maxy-miny); //      cout<<take<<" "<<t[1][take]<<endl;
    }
    for(int kk=2;kk<=k;kk++)FOR(take,1<<n){
      t[kk][take]=t[kk-1][take];
      for(int take2=take;take2;take2=(take2-1)&take)
        t[kk][take]<?=t[kk-1][take-take2]>?t[1][take2];
    }
    cout<<"Case #"<<c<<": "<<t[k][(1<<n)-1]<<endl;   }   return 0; }
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3 answers

EDIT:

Its extension in older versions of GCC. See this:

> ? =?

<? >?. <?=.

6.1 ++


EARLIER POST:

, ++ . , , .

, , <=, ?:

, , , ( , , ).

it doesn't compile on my system. ? .

+4

, , . , .

+1

Is it possible that you are mixing with this statement: a? BC. Otherwise, please post a link to some code in which you saw it, because it is sure that this is not normal with C ++. It may be a different language, but googling it literally returns nothing, so I don’t think the answer is.

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