conversion from __int64 to unsigned int

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unsigned int milliseconds = (unsigned int)(r & 0xffffffff);

What is the meaning of above statement(here r is __int64)? Please explain

Posted 14-Sep-10 3:01am

project.gearup

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Young J. PARK · Answer 1 · 2010-09-20T01:13:00

__int64 r;
unsigned int milliseconds_1;
unsigned int milliseconds_2;
unsigned int milliseconds_3;

r = 0x123456789;
milliseconds_1 = (unsigned int)(r & 0xffffffff); //#1
milliseconds_2 = (unsigned int)r; //#2
milliseconds_3 = r; //#3

Actually if you run this code in visual studio, will works without any error or warning.

#1 and #2 is exactly same.
But #3 is implicit usage. Some compiler will complain because assign large variable(64 bit) value to small variable(32bit).

HimanshuJoshi · Answer 2 · 2010-09-14T03:08:00

Solution 1

It masks off the 32 bits from the 64 bit integer, preserves only lower 32 bits of value and assignes it to a 32 bit integer.

Posted 14-Sep-10 3:08am

HimanshuJoshi

Updated 14-Sep-10 3:22am

v2

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project.gearup 14-Sep-10 9:18am

can you please elaborate your answer? cos lower 32 means interms of place or value?

HimanshuJoshi 14-Sep-10 9:21am

In terms of value

Emilio Garavaglia · Answer 3 · 2010-09-20T01:38:00

0xFFFFFFFF in binary are just 32 1s.

if r is an __int64, the constant (that is an integer) is promoted to __int64 itself, becoming 0x00000000FFFFFFFF (32 zeroes followed by 32 1s)

The bitwise AND (&) will result as 32 zeroes followed by the 32 rightmost bits of r.

The cast into (unsigned int) will cut away the 32 leftmost bits, giving a value that is assignable to milliseconds.

The conversion is not properly "safe", since it suffer of the mixing between signed and unsigned values.
(if r is -1, milliseconds will be 2³²-1)

A one-shot cast ( milliseconds = (unsigned int)r ) would had been the same.

conversion from __int64 to unsigned int

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