Interpret 4 char as an integer

Question

I have a function that returns n random bytes. For testing I wanted it to produce 4 random bytes and then assign the 4 bytes of an integer as the random bytes. I have something that works but it seems like an excessive amount of syntax for what I am trying to do. I just wanted to see if it really is necessary or if it can be done in a better way

Note: This is being done in xv6, no solutions that involve an include

int val0 = (((int)buf[0])&0xff);
int val1 = (((int)buf[1])&0xff);
int val2 = (((int)buf[2])&0xff);
int val3 = (((int)buf[3])&0xff);

int number = val0 | (val1<<8) | (val2<<16) | (val3<<24);

Note this is swapping the endianness. buf[0] goes from the most significant byte to the least. — D'Nabre, Nov 09 '14 at 03:55
@D'Nabre, Didn't notice, the bytes are random, but good point — asimes, Nov 09 '14 at 03:58
@D'Nabre: on some machines, it would not be swapping the endianness. — Jonathan Leffler, Nov 09 '14 at 04:10
`buf[i]` will always be promoted to `int` so you don't need the cast. If buf is unsigned you don't need to mask either. Better use `val0 = (unsigned char)buf[0];` — phuclv, Nov 11 '14 at 12:26

Shafik Yaghmour · Accepted Answer · 2014-11-11T11:19:03.713

4

As far as I can see you are using gcc from this reference here and we know gcc support type punning through unions explicitly then that may be a reasonable solution:

union charToInt
{
    char arr[sizeof(int)] ;
    int x ;
} ;

int main()
{
    union charToInt u1 = { .arr = "ab" } ; // or { .arr = "abcd" } if sizeof(int) == 4
}

Then u1.x will hold the the int value. I did not add any output since I am not sure what you use for output.

edited Nov 11 '14 at 11:19

answered Nov 09 '14 at 03:52

Shafik Yaghmour

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Great, thank you. I never had a good reason to use unions before – asimes Nov 09 '14 at 03:56
@asimes the other main use would be to implement a variant type. – Shafik Yaghmour Nov 09 '14 at 04:02

Interpret 4 char as an integer

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