what does this C++ line of code mean "sol<?=f((1<<n)-1,i,0)+abs(P[i])*price;"

Question

Could anyone help me to understand following line of code:

sol< ?=f((1<< n)-1,i,0)+abs(P[i])*price;

I am studying an algorithm written using c++ and it has following operator < ?= . My problem is with understanding < ?= operator. Also when I compile this code using g++ compiler , it gives error message for above line of code line of code

following is the error message returned.

Hello.cpp: In function ‘int main()’:

Hello.cpp:115: error: ‘memset’ was not declared in this scope

Hello.cpp:142: error: expected primary-expression before ‘?’ token

Hello.cpp:142: error: expected primary-expression before ‘=’ token

Hello.cpp:142: error: expected ‘:’ before ‘;’ token

Hello.cpp:142: error: expected primary-expression before ‘;’ token

Maybe < ?= it is not a single operator, but I can not understand what exactly this line of code does.

Thanks in advance for the time you spent reading this post.

Answer 1

This line isn't a line of code. That's why it doesn't compile. It is meaningless to ask what it does.

Answer 2

Take a look at the C grammar here

The only use of ? is in the ternary operator:

conditional_expression
    : logical_or_expression
    | logical_or_expression '?' expression ':' conditional_expression
    ;

Where the ? is followed by an expression. This does not happen in your case. So your code is not a valid C.

Answer 3

It could almost be a line of PHP code though: all it needs is remove a space to form at the end.

<?= foo(); ?>

is equivalent to

<?php echo foo(); ?>

Answer 4

It's a GNU extension. It's basically a "lower than" operator.

int a = 3;
a <?= 2;
cout << a << endl; // prints 2, because 2 < 3

a <?= 10;
cout << a << endl; // prints 2 as well, because 10 > 2

Read more here.

Answer 5

To be clear for anyone reading this and not being able to follow; <?= and >?= are assignment versions of <? and >? which are depricated GCC extensions which served the purpose of (x>y)?x:y or (x<y)?x:y respectively.

Therefore, x <?= y; would be x = x <? y; which is x = (x<y) ? x : y;

Most compiler vendors introduce language extensions, and many make it into future language standards. Usually these extensions are either just very easy to add or make the job of writing standard libraries much easier.