int a = 1; // 0001b
int b = a << 1; // 0010b
std::cout << "a = " << a << ", b = " << b << std::endl;
Output
a = 1, b = 2
Why
The left bit wise shift will shift the bits of the left hand value (a
) the number specified on the right (1
), essentially padding the least significant bits with 0's, so shifting the value of 5
(binary 0000 0101
) to the left 4 times (e.g. 5 << 4
) will yield the value of 80
(binary 0101 0000
). You might note that shifting a value to the left 1 time is also the same as multiplying the value by 2, example:
int a = 7;
while (a < 200) {
std::cout << "a = " << a << std::endl;
a <<= 1;
}
a = 7;
while (a < 200) {
std::cout << "a = " << a << std::endl;
a *= 2;
}
But it should be noted that the left shift operation will shift all bits to the left, including the sign bit, example:
int a = 2147483647; // 0111 1111 1111 1111 1111 1111 1111 1111
int b = a << 1; // 1111 1111 1111 1111 1111 1111 1111 1110
std::cout << "a = " << a << ", b = " << b << std::endl;
Possible output: a = 2147483647, b = -2
While some compilers will yield results that seem expected, it should be noted that if you left shift a signed number so that the sign bit is affected, the result is undefined. It is also undefined if the number of bits you wish to shift by is a negative number or is larger than the number of bits the type on the left can hold, example:
int a = 1;
int b = a << -1; // undefined behavior
char c = a << 20; // undefined behavior
The bit wise left shift does not change the value of the original values unless specifically assigned to using the bit wise assignment compound operator <<=
:
int a = 5; // 0101b
a <<= 1; // a = a << 1;