Is it possible to store 2 32-bit values in one long int variable?

Question

I want to store two 32-bit values in a single long int variable.

How would you do this on a 32-bit OS using C? Is it possible to store the data in a single long long variable? If so, how is that done?

code is already present, so I just want to know 64bit data can be store in one single long int variable or not — Syedsma, Aug 11 '11 at 14:22
@Syedsma: In addition to the given answers it may interest you, that on Windows the [MAKELPARAM](http://msdn.microsoft.com/en-us/library/ms632661\(v=vs.85\).aspx) macro does this (except it combines 2x16-bit into one 32-bit word). [LOWORD](http://msdn.microsoft.com/en-us/library/ms632659\(v=vs.85\).aspx), [HIWORD](http://msdn.microsoft.com/en-us/library/ms632657\(v=vs.85\).aspx) can then later be used to extract the high and low words. — user786653, Aug 11 '11 at 14:27

cnicutar · Answer 1 · 2011-08-11T14:27:22.083

10

Use an uint64_t and bitwise operators.

uint64_t i64;
uint32_t a32, b32;

// Be carefull when shifting the a32.
// It must be converted to a 64 bit value or you will loose the bits
// during the shift. 
i64 = ((uint64_t)a32 << 32) | b32;

edited Aug 11 '11 at 14:27

answered Aug 11 '11 at 14:19

cnicutar

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after that can i store i64 in long int variable – Syedsma Aug 11 '11 at 14:21
@Syedsma I think you can store it in an `unsigned long long`. – cnicutar Aug 11 '11 at 14:22
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@Martin Thanks! I always forget that :-) I changed the cast to a C-style one. – cnicutar Aug 11 '11 at 14:23
Failure to cast before the shift will not just loose bits. It will invoke undefined behavior because the shift amount is greater than or equal to the width of the type. – R.. GitHub STOP HELPING ICE Aug 11 '11 at 14:37
@Syedsma A `long` might be too small. – cnicutar Aug 11 '11 at 14:45
@Syedsma: an `unsigned long int` may be 32 bits, so use `#include `and the types `uint64_t` and `uint32_t`, as shown. They do have the required size indicated by their names. How they map to `long` or `long long` or whatever is compiler and platform dependent, and can be found in `stdint.h` – Rudy Velthuis Aug 11 '11 at 21:36

score 0 · Accepted Answer · answered Aug 11 '11 at 14:19

0

Assuming a long is 64 bits on your platform,

int v1 = 123;
int v2 = 456;
long val = v1 << 32 | v2;

answered Aug 11 '11 at 14:19

Kevin

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long may only be 32 bits. Then val == v2 – Martin York Aug 11 '11 at 14:20
"assuming a long is 64 bits on your platform". If you're running on a 64 bit box with a 64 bit compiler, a long is 64 bits. – Kevin Aug 11 '11 at 14:39
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"warning: left shift count >= width of type" - you need a cast in ther e somewhere. Also try it with `v2 = -1` and watch it blow up... – user786653 Aug 11 '11 at 14:50

score 0 · Answer 3 · answered Aug 11 '11 at 14:22

0

Unless sizeof(long int) == 8, the answer is no. If that equality is true, then use Kevin's or cnicutar's answer.

answered Aug 11 '11 at 14:22

Bill Lynch

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Not quite correct. The standard allows `INT_MIN` to be -32767 and `INT_MAX` to be 32767. So what is really required is for the type used to store the result to be twice the size of the operands (original values) at least. – jweyrich Aug 11 '11 at 14:35
@jweyrich: I understand the standard allows an int to be 16 bits in size. However, he asked if he could fit two 32bit numbers into a long int. If a long int is 4 bytes long, that would be quite difficult. Also, a long int simply must be greater or equal to the size of an int. – Bill Lynch Aug 11 '11 at 16:10
My point is that your answer is assuming that `int` is 32 bits, which might not be the case. – jweyrich Aug 11 '11 at 16:41

Is it possible to store 2 32-bit values in one long int variable?

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