Why does shifting more than the allowed bits still work?

Question

I have an int8_t and I wanted to see what would happen if I shift it left further than 8 bits. So this is what I did:

int8_t x = 1;

std::cout << (x << 10);

For some reason this returns 1024 as if the type contained enough bits to represent that number. I thought that when you shift more than the given bits you would get 0 in all the bits (or signed overflow/underflow which leads to undefined behavior). Also, I ran this code to give me the maximum number of int8_t:

std::numeric_limits<int8_t>::max(); // 127

The max number of this type is 127 but shifting it left can make it even go higher than its unsigned type! How is this possible?

NPE · Accepted Answer · 2013-03-24T15:15:26.203

9

The arguments to << are being implicitly widened to int, and the result of x << 10 is also an int.

edited Mar 24 '13 at 15:15

answered Mar 24 '13 at 15:08

NPE

486,780
108
951
1,012

How can I keep it to `int8_t`? – user2030677 Mar 24 '13 at 15:08
1

@user2030677 You can't. You can cast the result after the shift,if you want. Then you get the implementation-defined conversion of an out-of-range value to a signed integer type. – Daniel Fischer Mar 24 '13 at 15:10
@DanielFischer ans NPE: Suppose if I shift like: `int8_t y = 10; std::cout << (x << y);` then it will return int **?** or output would be `0` – Grijesh Chauhan Mar 24 '13 at 15:10
@GrijeshChauhan You didn't tell us the value of `x` and how many bits are in your `int`. – Alexey Frunze Mar 24 '13 at 15:14
1

@GrijeshChauhan The operands of the shift operators are subject to the integer promotions, so you always get at least `int`. – Daniel Fischer Mar 24 '13 at 15:15
@AlexeyFrunze `x` is `int8_t x = 1;` from OP – Grijesh Chauhan Mar 24 '13 at 15:16
1

@GrijeshChauhan In that case, there's nothing new, you'll get 1024 printed. The type on the right side of `<<` and `>>` does not influence the type of the result. – Alexey Frunze Mar 24 '13 at 15:17
Is there any point in making them `int8_t` in the first place? – David G Mar 24 '13 at 15:33

Why does shifting more than the allowed bits still work?

1 Answers1