CODEWITHSUNDEEP
c++gccx86-64disassembly
galinette
galinette

Reputation: 9292

Strange bit operation result in C++ / disassembly?

In a C++ function, I have the following variables:

uint32_t buf = 229;  //Member variable
int bufSize = 0;     //Member variable
static constexpr const uint32_t all1 = ~((uint32_t)0);

And this line of code:

uint32_t result = buf & (all1 >> (32-bufSize ));

Then the value of result is 229, both by console output, or by debugging with gdb. The expected value is of course 0, and any trial to make a minimal example reproducing the problem failed.

So I went to the disassembly and executed it step by step. The bit shift operation is here:

0x44c89e  <+0x00b4>        41 d3 e8                 shr    %cl,%r8d

The debugger registers show that before the instruction, rcx=0x20 and r8=0xFFFFFFFF

After the instruction, we still have r8=0xFFFFFFFF

I have a very poor knowledge of x86-64 assembly, but this instruction is supposed to be an unsigned shift, so why the heck isn't the result 0?

I am using mingw-gcc x86-64 4.9.1

Upvotes: 0

Views: 121

Answers (1)

shiv
shiv

Reputation: 1952

Invoke -Wall on compiler and you will have -Wshift-count-overflow problem as you are using 32 for shifting which is size of unsigned int. Now you can do one thing just for knowing about it. Change 32 to 31 and compile. Then compare the assembly generated and you will know what went wrong.

The easiest fix for you would be use long data type for all1 and result.

Upvotes: 2

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