Bitwise shift operation in C on uint64_t variable

Question

I have the following sample code:

uint64_t x, y;
x = ~(0xF<<24);
y = ~(0xFF<<24);

The result would be:

x=0xfffffffff0ffffff
y=0xfffff

Can anyone explain the difference? Why x is calculated over 64 bit and y only on 32?

Answer 1

The default operation is 32 bit.

x=~(0xf<<24);

This code could be disassembled into the following steps:

int32_t a;
a=0x0000000f;
a<<=24;   // a=0x0f000000;
a=~a;     // a=0xf0ffffff;
x=(uint64_t)a;  // x = 0xfffffffff0ffffff;

And,

y = ~(0xFF<<24);

int32_t a;
a=0x000000ff;
a<<=24;   // a=0xff000000;
a=~a;     // a=0x00ffffff;
x=(uint64_t)a;  // x = 0x000000000ffffff;

Answer 2

Because 0x0f << 24 is a positive number when viewed as an int , it's sign-extended to a positive number, ie to 0x00000000_0f000000 (the underscore is just for readability, C does not support this syntax). This is then inverted into what you're seeing.

0xff << 24 on the other hand is negative, so it's sign-extended differently.

Answer 3

Other posters have shown why it does this. But to get the expected results:

uint64_t x, y; 
x = ~(0xFULL<<24); 
y = ~(0xFFULL<<24);

Or you can do this (I don't know if this is is any slower than the above though):

uint64_t x, y; 
x = ~(uint64_t(0xF)<<24); 
y = ~(uint64_t(0xFF)<<24); 

Then:

x = 0xfffffffff0ffffff
y = 0xffffffff00ffffff

Answer 4

You have undefined behavior in your program so anything might happen.

• The integer literals 0xF or 0xFF are of type int , which is equivalent to signed int . On this particular platform, int is apparently 32 bits.
• The integer literal 24 is also a (signed) int .
• When the compiler evaluates the << operation, both operands are (signed) int so no implicit type promotions take place. The result of the << operation is therefore also a (signed) int .
• The value 0xF<<24 = 0x0F000000 fits in a (signed) int as a non-negative value, so everything is ok.
• The value 0xFF<<24 = 0xFF000000 does not fit in (signed) int ! Here, undefined behavior is invoked and anything might happen.

ISO 9899:2011 6.5.7/4:

"The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros." /--/

"If E1 has a signed type and nonnegative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

So the expression 0xFF<<24 can't be used. The program is free to print any garbage value after that.

But if we ignore that one and focus on 0x0F<24:

• 0x0F000000 is still a (signed) int . The ~operator is applied to this.
• The result is 0xF0FFFFFF, which is still a signed int. And on almost any system, this 32-bit hex equals a negative number in two's complement.
• This signed int is converted to the type uint64_t during assignment. This is done in two steps, first by converting it to a signed 64 bit, then by converting that signed 64 to an unsigned 64.

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Bugs like this is why the coding standard MISRA-C contains a number of rules to ban sloppy use of integer literals in expression like this. MISRA-C compliant code must use the u suffix after each integer literal (MISRA-C:2004 10.6) and the code is not allowed to perform bitwise operations on signed integers (MISRA-C:2004 12.7).