bitshift in matlab vs ishft in fortran

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!

[Shashi Kiran]

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?

Answer 1

[Steven Lord]

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

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format hex
x

x = 1×2

03625cbe b5072949

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y

y = int64
   b507294903625cbe

The swapbytes function may also be of interest to you.