Anonymous, or unknown.
An unsigned artist.
Unsigned? Not much.
The Unsigned Guide was created in 2003.
Having an unsigned integer means that the integer is positive, and not negative; literally, the integer is unsigned and assumed to be positive. The unsigned integer 8 is positive-eight, not negative-eight.
#include<iostream> #include<vector> unsigned count_digits (unsigned num, const unsigned base=10) { unsigned count=1; while (num/=base) ++count; return count; } class number { std::vector<unsigned> value; unsigned base; public: number (const unsigned _value, const unsigned _base=10): value {}, base {_base} { *this = _value; } number& operator= (const unsigned _value); operator unsigned () const; bool is_narcissistic () const; }; number& number::operator= (unsigned _value) { unsigned count = count_digits (_value, base); value.resize (count); while (count) { value[value.size()-count--] = _value%base; _value/=base; } return *this; } number::operator unsigned () const { unsigned num = 0; for (unsigned index=0; index<value.size(); ++index) num += value[index]*static_cast<unsigned>(std::pow (base, index)); return num; } bool number::is_narcissistic () const { unsigned num = 0; for (unsigned index=0; index<value.size(); ++index) num += static_cast<unsigned>(std::pow (value[index], value.size())); return num == static_cast<unsigned> (*this); } unsigned main() { const unsigned min=1; const unsigned max=100; std::cout << "Narcissistic numbers in the range " << min << " through " << max << ":\n\t"; for (unsigned n=min; n<=max; ++n) if (number(n).is_narcissistic()) std::cout << n << ' '; std::cout << '\n' << std::endl; }
No. Java uses no unsigned numbers.
for example: unsigned char attach (unsigned char byte, unsigned char bit) { unsigned char mybyte; mybyte = byte&0x7f; if (bit) mybyte |= 0x80; return mybyte; }
If you assign -1 to a unsigned variable it will contain the biggest number its able to hold. For example if you assign -1 to a unsigned int it will be 4294967295 as its the biggest number a unsigned int can hold.
#include<stdio.h> unsigned sum_row (unsigned* sq, const unsigned width, const unsigned row) { unsigned sum, col; sum = 0; for (col=0; col<width; ++col) sum += sq[row*width+col]; return sum; } unsigned sum_col (unsigned* sq, const unsigned width, const unsigned col) { unsigned sum, row; sum = 0; for (row=0; row<width; ++row) sum += sq[row*width+col]; return sum; } unsigned sum_diag (unsigned* sq, const unsigned width) { unsigned sum, row, col; sum = 0; for (row=0, col=0; row<width; ++row, ++col) sum += sq[row*width+col]; return sum; } unsigned sum_anti (unsigned* sq, const unsigned width) { unsigned sum, row, col; sum = 0; for (row=0, col=width-1; row<width; ++row, --col) sum += sq[row*width+col]; return sum; } bool is_magic (unsigned* sq, const unsigned width) { unsigned magic, row, col; magic = sum_row (sq, width, 0); for (row=1; row<width; ++row) if (magic!=sum_row(sq, width, row)) return false; for (col=0; col<width; ++col) if (magic!=sum_col(sq, width, col)) return false; if (magic!=sum_diag(sq, width)) return false; if (magic!=sum_anti(sq, width)) return false; return true; } int main () { const unsigned width = 3; unsigned a[width][width] {{2,7,6},{9,5,1},{4,3,8}}; unsigned row, col; printf ("Square:\n\n"); for (row=0; row<width; ++row) { for (col=0; col<width; ++col) { printf ("%d ", a[row][col]); } printf ("\n"); } printf ("\n"); if (is_magic((unsigned*)&a, width)) printf ("The square is magic with a magic constant of %d\n", sum_row((unsigned*)&a, 3,0)); else printf ("The square is not magic\n"); return 0; }
No. They are unsigned, therefore all representations are positive.
#include<iostream> unsigned sum_of_digits(unsigned num) { unsigned sum = 0; do { sum += num%10; } while(num/=10); return sum } int main() { unsigned num = 42; unsigned sum = sum_of_digits (num); std::cout << sum; // output: 6 }
What is the significance of declaring a constant unsigned integer?