Bitmasks not working on 64-bit values.

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So i have these functions, which works perfectly:

C++

void pushw(int16_t x) {
	push((x & 0xFF00) >> 8);
	push(x & 0x00FF);
}

void pushd(int32_t x) {
	pushw((x & 0xFFFF0000) >> 16);
	pushw(x & 0x0000FFFF);
}

But when i do this:

C++

void pushq(int64_t x) {
	pushq((x & 0xFFFFFFFF00000000) >> 32);
	pushq(x &  0x00000000FFFFFFFF);
}

It compiles but it skips all the prints i have in the main function, from where i am also calling the push functions. Why is this?

What I have tried:

I have not tried anything, i am just utterly confused why it won't listen to my input.

Posted 25-Feb-23 9:13am

Jamie Engel

Updated 12-Mar-23 12:55pm

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k5054 25-Feb-23 15:26pm

should pushq call pushd, or is that a typo?

Jamie Engel 26-Feb-23 12:39pm

yes it was a typo, it wasn't meant to be infinitely recursive haha. I don't know how to delete my question though, seems kind of unnecessary to take up bandwidth just because i was tired and forgot to check the spelling xD

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OriginalGriff · Answer 1 · 2023-02-25T09:30:00

Look at your code:

void pushq(int64_t x) {
	pushq((x & 0xFFFFFFFF00000000) >> 32);
	pushq(x &  0x00000000FFFFFFFF);
}

So pushq calls pushq, which calls pushq, which ... continues until out of stack space and your app crashes.
Did you mean to call pushd instead? If so, you probably want to cast the values as well to avoid warnings on some compilers.

hans.sch · Answer 2 · 2023-03-12T06:23:00

The constant 0xFFFFFFFF00000000 is truncated to 32 bits, yielding 0. You should append "ll" to indicate that the constant is 64 bits wide: 0xFFFFFFFF00000000ll. Note that this is two times lower case L, and not number eleven or the logical OR operator.
An alternative solution is to just call pushd(x>>32).

Patrice T · Answer 3 · 2023-03-12T12:55:00

Quote:
Bitmasks not working on 64-bit values.

C++ compiler use a default datatype if 32 bits if it is not told otherwise. There is no type inference, you are responcible of telling the compiler which datatype to use.
Below links speak mainly about cast on variables, but it also apply to constants in formulas:
C++ Casting Operators[^]
Const cast in C[^]
Explicit type conversion - cppreference.com[^]
So:

C++

void pushq(int64_t x) {
	pushq((x & 0xFFFFFFFF00000000) >> 32);
	pushq(x &  0x00000000FFFFFFFF);
}

and

C++

void pushq(int64_t x) {
	pushq((x & (ll)0xFFFFFFFF00000000) >> 32);
	pushq(x &  (ll)0x00000000FFFFFFFF);
}

gives different results.

Jamie Engel · Answer 4 · 2023-02-26T06:40:00

Solution 2

There was a misspelling. It was meant to call the function pushd, not itself (pushq).

Posted 26-Feb-23 6:40am

Jamie Engel

Bitmasks not working on 64-bit values.

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