C++ – How compare vector size with an integer?

cvector

I am using the following code to throw an error if the size of vector (declared as vector<int> vectorX) is is different than intended.

vector<int> vectorX;
int intendedSize = 10;
// Some stuff here
if((int)(vectorX.size()) != (intendedSize)) {
    cout << "\n Error! mismatch between vectorX "<<vectorX.size()<<" and intendedSize "<<intendedSize;
    exit(1);
}

The cout statement shows the same size for both. The comparison is not showing them to be equal.

Output is Error! mismatch between vectorX 10 and intendedSize 10

Where is the error? Earlier I tried (unsigned int)(intendedSize) but that too showed them unequal.

Best Answer

I'm writing this answer because the other two, including the accepted one, are both wrong. The type of std::vector's size() is not unsigned int, nor it is size_t.

The type of the size of an std::vector<T> is std::vector<T>::size_type.

That's it. On some architecture and for some compilers it might be the same as size_t, in some others it might not. The assumption that a variable of type size_t can hold the same values than one of type std::vector<T>::size_type can fail.

To check that your vector has the right size you could do something like:

if(vec.size() != static_cast<std::vector<int>::size_type>(expected_size)) {
    std::cerr << "Error!" << std::endl;
}

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", ':');