bitshift in matlab vs ishft in fortran

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Jaden Hoechstetter
Jaden Hoechstetter 2024-9-16
回答： Steven Lord 2024-9-16
Hi,
I am trying to convert a large 64 bit number into 2x 32 bit numbers, Here is my code:
U = bitshift(v, -24);
L = bitand(v, 0x000000ffffffs64);
This is replicated off of fortran code:
U = ishft(v, -24)
L = iand(v, Z'000000ffffff')
For value v = 2830037, the L's agree, but "bitshift(2830037, -24) = 0" and "ishft(2830037, -24) = 1".
I am confused. Any help would be appreciated!
  1 个评论
Shashi Kiran
Shashi Kiran 2024-9-16
Hi Jaden,
I see that U is zero when running the Fortran code(in onlinegdb compiler) as well. Are there any other ways to replicate the issue you are experiencing?

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回答（1 个）

Steven Lord
Steven Lord 2024-9-16
Ran in:
Instead of using bit operations yourself, why not just use typecast?
t = 'int32';
x = randi([intmin(t), intmax(t)], 1, 2, t) % 2 random int32 values
x = 1×2
56777918 -1257821879
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y = typecast(x, 'int64')
y = int64 -5402303834441491266
z = typecast(y, t)
z = 1×2
56777918 -1257821879
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
format hex
x
x = 1×2
03625cbe b5072949
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
y
y = int64
b507294903625cbe
The swapbytes function may also be of interest to you.

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