Convert unsigned char array with int to unsigned char

Question

Can somebody help me out with below error. Shall I cast len before I try to pass buf?

int len=2;
unsigned char tmp[len + 1];
unsigned char * buf = &tmp;

The error is:

error: cannot convert 'unsigned char (*)[(((unsigned int)((int)len)) + 1)]' to 'unsigned char*' in assignment

array internal representation is as a pointer. That is in your case type of tmp is unsigned char * and type of buf is unsigned char *.So type miss match you try to assign char ** to char *. — rajesh6115
– rajesh6115, Commented Dec 13, 2012 at 11:55
@Rajesh: Arrays are not pointers. The type of tmp is unsigned char[3]. Nothing else. There is no unsigned char** here anywhere, and certainly no char**. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Dec 13, 2012 at 12:01
@Lightness Races in Orbit:sorry for forgetting to add constant pointer (pointer value can not be changed) which points to a fixed memory location only. But I explain it only for type miss match Nothing else. — rajesh6115
– rajesh6115, Commented Dec 13, 2012 at 12:10
@Rajesh: The type mismatch is between unsigned char* (pointer to an unsigned char) and unsigned char(*)[3] (pointer to an array of three unsigned chars), not between unsigned char* and unsigned char**. Note that I do ignore the VLA component and pretend that the array dimension is fixed at 3, as it's supposed to be. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Dec 13, 2012 at 12:12
@LightnessRacesinOrbit : I don't for c++ I say all above things(type mismatch according to error highlighted ) . And error come only for the use of '&' address of operator only. This discussion can go to a bit longer but user may divert and confuse also. and I provide the simplest answer to the problem. You can ans this in a longer way. — rajesh6115
– rajesh6115, Commented Dec 13, 2012 at 12:28

Community · Accepted Answer · 2017-05-23 10:32:17Z

3

If you just want a pointer to the array, use

unsigned char * buf = tmp;

By the way, the way you declare tmp makes it a variable-length array (VLA). Technically, these are not allowed in C++, although many compilers support VLAs as an extension.

edited May 23, 2017 at 10:32

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answered Dec 13, 2012 at 11:55

NPE

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1 Comment

davidnr Over a year ago

Also to allow it to compile, len must be declared as const.

Lightness Races in Orbit · Accepted Answer · 2012-12-13 11:59:21Z

2

The problem is not len. The problem is that you are trying to take the address of the array, rather than the address of the first element of the array; the types simply don't match.

Instead:

unsigned char* buf = &tmp[0];

Also, the array name conveniently (FSVO "conveniently") decays to &tmp[0] anyway, so you could write simply:

unsigned char* buf = tmp;

In addition you are presently not using a constant value for the array's dimension — ensure that len is made const and initialised with a constant value. In C++11, this'd better be constexpr to really ensure that you're not accidentally attempting to use GCC VLAs.

answered Dec 13, 2012 at 11:59

Lightness Races in Orbit

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5 Comments

Steve Jessop Over a year ago

Ooh, what case does C++11 break so that you have to add constexpr rather than just having a const int with a visible initializer that's an ICE?

Lightness Races in Orbit Over a year ago

@SteveJessop: I'm not aware of any, but surely constexpr is preferable by its nature? If you always ensure you have a visible initializer then great, but asking the OP to remember to do this seems like a bit of a reach. No offence, OP.

Steve Jessop Over a year ago

ah, I was concentrating on "really ensure" rather than "better". I thought you meant that const int in C++11 doesn't really ensure that. AFAIK you're right that constexpr is a better way of expressing it in C++11.

Lightness Races in Orbit Over a year ago

@SteveJessop: I meant rather that using const int -- and just hoping that you remembered everything else you need -- doesn't ensure that. ;)

Steve Jessop Over a year ago

Yes, it makes a bigger difference with const int a = foo(); where foo() may or may not be a constexpr function. constexpr int a = foo(); checks that it is. With const int a = 4; I'm reasonably confident I can verify by eye that 4 is a constant ;-)

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