How to uint32 to unsigned char array

Question

I have 8 uint32 elements and i want to break each uint32 into 4 uint8 then add all uint8 beside each others as unsigned chars in the array , how can i do that ?

possible duplicate of [Converting a UINT32 value into a UINT8 array\[4\]](http://stackoverflow.com/questions/6499183/converting-a-uint32-value-into-a-uint8-array4) — devnull, Jun 20 '13 at 05:19

score 1 · Answer 1 · answered Jun 20 '13 at 05:23

1

UINT32 value;
UINT8 byteval[4];

for(int i = 0 < 4; i++)
    byteval[i] = value >> (i*8);

answered Jun 20 '13 at 05:23

Devolus

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score 1 · Accepted Answer · answered Jun 20 '13 at 05:27

1

You can make use of the power of union for this

union value
{
   uint32 number;

   struct bytes
   {
       uint8 bytevalue[4];
   };
};

answered Jun 20 '13 at 05:27

hazzelnuttie

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2

I agree with using a union, but that inner struct is useless cruft. You can just write `union value { uint32_t quad; uint8_t bytes[4]; };`. – Will Jun 20 '13 at 06:00
1

The downside with union is that the code turns endian-dependent, you can't specify the order of the 4 individual bytes. If portability is important, then you can't use union. – Lundin Jun 20 '13 at 06:17

Kapil Chittewan · Answer 3 · 2013-06-20T08:48:00.710

0

Use Structure & Union in combination.

    typedef struct
{
    uint32 ArrayOf32Bit[8];
}Arrayof32bitVar_t;

typedef union
{
    Arrayof32bitVar_t Var8int32;
    uint8             Array8char[8*4]; // instead use macro
}tydefUnion_t;

func_add
{
    int i
    tydefUnion_t a; // 

    /*Here update variable a.Var8int32.ArrayOf32Bit*/

    int addResult = 0; 
    for(i;i<(8*4);i++)
    {
        addResult += a.Array8char[i];
    }
}

edited Jun 20 '13 at 08:48

answered Jun 20 '13 at 06:24

Kapil Chittewan

99
3

your second typedef in incomplete. – urzeit Jun 20 '13 at 07:57

How to uint32 to unsigned char array

3 Answers3