Left shift operation on an unsigned 8 bit integer

Question

I am trying to understand shift operators in C/C++, but they are giving me a tough time.

I have an unsigned 8-bit integer initialized to a value, for the example, say 1.

uint8_t x = 1;

From my understanding, it is represented in the memory like |0|0|0|0|0||0||0||1|. Now, when I am trying to left shit the variable x by 16 bit, I am hoping to get output 0. But to my surprise, I am getting 65536. I am certainly missing something which I am not able to get.

Here's my code:

#include <iostream>

int main() {
    uint8_t x = 1;
    std::cout<<(x<<16)<<"\n";
    return 0;
}

It's a naive question, but it is bothering me a lot.

Vlad from Moscow · Accepted Answer · 2020-02-07T16:44:39.260

6

In this expression

x<<16

the integer promotions are applied to the both operands. So the result of the expression is an object of the type int.

Try the following demonstrative program

#include <iostream>
#include <iomanip>
#include <type_traits>
#include <cstdint>


int main() 
{
    uint8_t x = 1;

    std::cout << std::boolalpha << std::is_same<int, decltype( x<<16 )>::value << '\b';

    return 0;
}

Its ouput is

true

From the C++ Standard (8.8 Shift operators)

... The operands shall be of integral or unscoped enumeration type and integral promotions are performed. The type of the result is that of the promoted left operand.

edited Feb 07 '20 at 16:44

answered Feb 07 '20 at 16:34

Vlad from Moscow

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Please correct me if I am wrong, the operand 16 is an integer of 32 bit(assuming an integer to be 32 bit), so, when applying it with x(or 8 bit), it promotes it also promotes it to 32 bit and finally, the result is of type int(32 bit). Am I right ? – Pankaj Mishra Feb 07 '20 at 16:41
@PankajMishra See my updated post. – Vlad from Moscow Feb 07 '20 at 16:45
Very clear,now. Thanks. Upvoted. – Pankaj Mishra Feb 08 '20 at 05:43
Strangely, if you have code like this: `unsigned char value = 255; unsigned char shiftAmount = 24; printf("%d ", (value << shiftAmount) );` you get -16777216 as the output (not 0) – bobobobo Sep 17 '21 at 01:47
@bobobobo The operands of the shift expression are promoted to the type int. So the sign bit is set. – Vlad from Moscow Sep 17 '21 at 03:24

Left shift operation on an unsigned 8 bit integer

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