Converting Char array to Long in C

Question

This question may looks silly, but please guide me I have a function to convert long data to char array

void ConvertLongToChar(char *pSrc, char *pDest)
{
    pDest[0] = pSrc[0];
    pDest[1] = pSrc[1];
    pDest[2] = pSrc[2];
    pDest[3] = pSrc[3];
}

And I call the above function like this

long lTemp = (long) (fRxPower * 1000);
ConvertLongToChar ((char *)&lTemp, pBuffer);

Which works fine. I need a similar function to reverse the procedure. Convert char array to long. I cannot use atol or similar functions.

Solution 1

Leaving the burden of matching the endianness with your other function to you, here's one way:

unsigned long int l = pdest[0] | (pdest[1] << 8) | (pdest[2] << 16) | (pdest[3] << 24);

Just to be safe, here's the corresponding other direction:

unsigned char pdest[4];
unsigned long int l;
pdest[0] = l         & 0xFF;
pdest[1] = (l >>  8) & 0xFF;
pdest[2] = (l >> 16) & 0xFF;
pdest[3] = (l >> 24) & 0xFF;

Going from char[4] to long and back is entirely reversible; going from long to char[4] and back is reversible for values up to 2^32-1.

Note that all this is only well-defined for unsigned types.

(My example is little endian if you read pdest from left to right.)

Addendum: I'm also assuming that CHAR_BIT == 8. In general, substitute multiples of 8 by multiples of CHAR_BIT in the code.

Solution 2

You can do:

union {
 unsigned char c[4];
 long l;
} conv;
conv.l = 0xABC;

and access c[0] c[1] c[2] c[3]. This is good as it wastes no memory and is very fast because there is no shifting or any assignment besides the initial one and it works both ways.

Solution 3

A simple way would be to use memcpy:

char * buffer = ...;
long l;
memcpy(&l, buff, sizeof(long));

That does not take endianness into account, however, so beware if you have to share data between multiple computers.

Solution 4

If you mean to treat sizeof (long) bytes memory as a single long, then you should do the below:

char char_arr[sizeof(long)];
long l;
memcpy (&l, char_arr, sizeof (long));

This thing can be done by pasting each bytes of the long using bit shifting ans pasting, like below.

l = 0;
l |= (char_arr[0]);
l |= (char_arr[1] << 8);
l |= (char_arr[2] << 16);
l |= (char_arr[3] << 24);

If you mean to convert "1234\0" string into 1234L then you should

l = strtol (char_arr, NULL, 10); /* to interpret the base as decimal */

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Comments

[josesuero]

It is also, according to the C++ standard, undefined behavior. :)

[Engineer]

@jalf This is a C question.

[josesuero]

@NickWiggill fair enough. It is also, according to the C standard, undefined behavior. :)

[Vinicius Kamakura]

this is not UB under gcc and MS compiler I am pretty sure. It probably is by the C std.

[underscore_d]

^ What compilers do isn't actually informative at all about what the Standard says, though you happened to be correct anyway. @jalf No, C made type-punning via a union well-specified in C99 onwards: stackoverflow.com/questions/11639947/… C++ did not inherit this, however.

[Antti Haapala -- Слава Україні]

and this is wrong, aliasing error and possible undefined behaviour from wrong alignment

[Antti Haapala -- Слава Україні]

and this is wrong, aliasing error and possible undefined behaviour from wrong alignment

[Beezer]

I tried your example but it did not work for me. What did, was something very similar: char* vIn = "0"; long vOut = strtol(vIn,NULL,10);

[Etienne de Martel]

@Beezer That does a very different thing: your method parses text into a number, while mine interprets the bytes of a buffer as a long. These are different problems.