for with one parameter

Question

Reading that spinlock and other multitasking stuff, I faced to this code:

#include <boost/range/algorithm.hpp>
#include <boost/atomic.hpp>
#include <boost/thread.hpp>
#include <iostream>
#include <vector>

class SpinLock
{
    boost::atomic_flag flag;
public:
    void lock()
    {
        while( flag.test_and_set(boost::memory_order_acquire) )
            ;
    }
    bool try_lock()
    {
        return !flag.test_and_set(boost::memory_order_acquire);
    }
    void unlock()
    {
        flag.clear(boost::memory_order_release);
    }
};

int main()
{
    using namespace std; using namespace boost;

    SpinLock lock;
    vector<thread> v;
    for(auto i = 0; i!=4; ++i)
        v.emplace_back([&lock, i]
        {
            for(auto j = 0; j!=16; ++j)
            {
                this_thread::yield();
                lock_guard<SpinLock> x(lock);
                cout << "Hello from " << i << flush << "\tj = " << j << endl;
            }
        });
    for(auto &t: v)
        t.join();
}

Could you explain why for has only one parameter?

And what that colon operator does?

And what is that t object?

score 3 · Accepted Answer · answered Oct 28 '14 at 06:55

3

It's range-based for. t has type std::thread&.

answered Oct 28 '14 at 06:55

ForEveR

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Laura Maftei · Answer 2 · 2014-10-28T07:06:05.240

2

It's a range-based for loop. These were introduced with C++11. You are iterating over a container.

Its syntax is:

for ( range_declaration : range_expression ) loop_statement

The for loop is executed from the beginning of the container to the end.

edited Oct 28 '14 at 07:06

answered Oct 28 '14 at 06:55

Laura Maftei

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score 1 · Answer 3 · answered Oct 28 '14 at 06:59

1

Example:

int  an_array[]={1,2,3,5,6,56,34,65,3,234};
for(int a : an_array)
   cout << a;

this for loop is equivalent to:

for(int index = 0; index < sizeof(an_array) / sizeof(an_array[0]) /*10*/; ++index)
   cout << an_array[index];

The supporting compilers do understand what vector is, if used against a range-based for loop, and they call begin and end for the container. Therefore, the last loop is same as:

for(vector<thread>::itereator t =  v.begin(); t != v.end(); ++t)
        t->join(); // Or. (*t).join()

answered Oct 28 '14 at 06:59

Ajay

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Do you mead that disassembly on regular x86 processor will give the same result in both cases? – Qeeet Oct 28 '14 at 07:03
1

Yes. The generated binary/assembly will be same. – Ajay Oct 28 '14 at 07:04

score 0 · Answer 4 · answered Oct 28 '14 at 07:00

0

range-based for loop, you dont need to know size, it means for all t's in container v do something (join in this case). somewhat diffrent from traditional for loop

answered Oct 28 '14 at 07:00

Ali Kazmi

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for with one parameter

4 Answers4