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Give a one-line C expression to test whether a number is a power of 2?
The one line expression is: ((0 != n) && !(n & n-1)) example: int main () { for (int n = 0; n <= 1000001; ++n) { if ((0 != n) && !(n & n-1)) cout << n << " is a power of 2" << endl; } return 0; } will produce: 1 is a power of 2 2 is a power of 2 4 is a power of 2 8 is a power of 2 16 is a power of 2 32 is a power of 2 64 is a power of 2 128 is a power of 2 256 is a power of 2 512 is a power of 2 1024 is a power of 2 2048 is a power of 2 4096 is a power of 2 8192 is a power of 2 16384 is a power of 2 32768 is a power of 2 65536 is a power of 2 131072 is a power of 2 262144 is a power of 2 524288 is a power of 2
What is 2 to the power of -2?
2 to the power of -2 is 0.25
What is 2 to the 2 power negative 3 to the 2 power plus 4 to the 2 power?
2 to the 2 power = 4 -3 to the 2 power = 9 4 to the 2 power = 16 You don't indicate what to do to the 4, but 9 + 16 = 25
What the value of 2 to the power of 1?
Just 2 2 to power 3 is 2x2x2=8 2 to power 2 is 2x2=4 2 to power 1 is 2
5 to the power of 2 plus 15 to the power of 2?
5 to the power of 2 plus 15 to the power of 2 = 250
