C++ – How to solve Unresolved inclusion: in eclipse

ceclipseeclipse-cdthelios

I built a simple program in eclipse:

#include <iostream>

using namespace std;
int main()
{
    cout << "TEST" << endl;
    return 0;
}

It worked in Visual Studio and CodeBlocks, but eclipse is acting weird. it says:

Unresolved inclusion: <iostream>

I read here:
C++ – Unresolved inclusion: <iostream>

and here: Unresolved <iostream> in Eclipse, Ubuntu

and neither of them worked.

Here are screenshots of project properties:

build_settings
general

edit:

I downloaded MinGW and now i have this under Settings:

Updated_Settings

How should i proceed?
Maybe now i don't need #include <iostream> because it's now included in the project?

include

I found iostream under Includes.

So i tried deleting #include <iostream>, but when i try to run the program i get:
Launch Failed. Binary not found. error:

launch_failed

Thanks

edit:

Seems like if i compile in some other program (say CodeBlocks) and create the exe file, then eclipse can run it. But it can't build its own exe.

Why?

Best Answer

This answer did not help me either. My issue was solved with the following steps:

You might try changing your source files from *.c to *.cpp. This will provoke gcc to think of the files as C++ and search the proper paths. Might need to make small modifications to the Makefile as well, like on the OBJ: line. Instead of:

OBJS = YourFile.o

try

OBJS = YourFile.cpp

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

C++ – How to iterate over the words of a string

I use this to split string by a delimiter. The first puts the results in a pre-constructed vector, the second returns a new vector.

#include <string>
#include <sstream>
#include <vector>
#include <iterator>

template <typename Out>
void split(const std::string &s, char delim, Out result) {
    std::istringstream iss(s);
    std::string item;
    while (std::getline(iss, item, delim)) {
        *result++ = item;
    }
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, std::back_inserter(elems));
    return elems;
}

Note that this solution does not skip empty tokens, so the following will find 4 items, one of which is empty:

std::vector<std::string> x = split("one:two::three", ':');