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I have this code in C (it's for study only):
char x;
uint64_t total = 0;
for(x = 20; x < 30; x++){
total = (((((1 << x) * x) / 64) + 1) * sizeof(uint64_t));
printf("%d - %llu\n", x, total);
}
What is printed:
20 - 2621448
21 - 5505032
22 - 11534344
23 - 24117256
24 - 50331656
25 - 104857608
26 - 218103816
27 - 18446744073625665544
28 - 18446744073575333896
29 - 18446744073508225032
Why at x > 26 do I have those strange values? I'm at gcc 4.6.1 on Ubuntu 10.10 64 bits.
485
<< is the left shift operator. It is shifting the number 1 to the left 0 bits, which is equivalent to the number 1 .
Left Shift and Right Shift Operators in C/C++ Takes two numbers, left shifts the bits of the first operand, the second operand decides the number of places to shift. Or in other words left shifting an integer “x” with an integer “y” denoted as '(x<<y)' is equivalent to multiplying x with 2^y (2 raised to power y).
Bitwise Shift Operators Left Shift(<<): The left shift operator, shifts all of the bits in value to the left a specified number of times. Syntax: value << num. Here num specifies the number of position to left-shift the value in value.
The bitwise shift operators are the right-shift operator ( >> ), which moves the bits of an integer or enumeration type expression to the right, and the left-shift operator ( << ), which moves the bits to the left.
Because 1 is an int, 32 bits, so (1 << 27)*27 overflows. Use 1ull.
Regarding your comment, if x is a uint64_t, then 1 << x is still an int, but for the multiplication it would be cast to uint64_t, so there'd be no overflow. However, if x >= 31, 1 << x would be undefined behaviour (as the resulting value cannot be represented by a signed 32 bit integer type).
158
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