Converting byte array to unsigned long in C++

Question

I am reading in binary data from a file:

char* buffIn = new char[8];
ifstream inFile(path, ifstream::binary);
inFile.read(buffIn, 8);

I then want to convert the char* read in (as binary) to an unsigned long but I am having problems - I am not quite sure what is going on, but for instance 0x00000000000ACD gets interpreted as 0xFFFFFFFFFFFFCD - I suspect all the 0x00 bytes are causing some sort of problem when converting from char* to unsigned long...

unsigned long number = *(buffIn);

How do I do this properly?

I wanted to suggest a solution but the question is already closed.:) — Vlad from Moscow
– Vlad from Moscow, Commented Feb 26, 2014 at 0:41

imreal · Accepted Answer · 2014-02-26 00:23:17Z

5

Since buffIn is of type char pointer, when you do *(buffIn) you are just grabbing one character. You have to reinterpret the memory address as an unsigned long pointer and then dereference it.

unsigned long number = *((unsigned long*)buffIn);

answered Feb 26, 2014 at 0:23

imreal

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9 Comments

Sundar Over a year ago

This according to the standards will be undefined behavior. char pointer need not have to be aligned to integer boundary.

adrianmcmenamin Over a year ago

Yes, I suppose that is obvious when you think about it. Thanks, will accept when I can.

imreal Over a year ago

@Sundar, it is aligned when pointing to allocated memory. Guaranteed by the standard. If the allocated memory is greater than unsigned longof course.

Sundar Over a year ago

@Nick start of the allocated memory (using malloc/calloc) is always aligned to the max possible pointer boundary (perhaps long long), but not the memory not allocated by malloc/calloc. securecoding.cert.org/confluence/display/seccode/…

imreal Over a year ago

@Sundar, from the standard:

1 Effects: The allocation function (3.7.3.1) called by a new-expression (5.3.4) to allocate size bytes of storage suitably aligned to represent any object of that size.

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Zac Howland · Accepted Answer · 2014-02-26 00:51:55Z

0

In addition to recasting the char[8] (which will only read the the first unsigned long - which is 32-bits in length), you can also use some simple bit-wise operations

unsigned long value = (((unsigned long)buffin[0]) << 24) | (((unsigned long)buffin[1]) << 16) | (((unsigned long)buffin[2]) << 8) | (unsigned long)buffin[3];

answered Feb 26, 2014 at 0:51

Zac Howland

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2 Comments

adrianmcmenamin Over a year ago

The unsigned longs are 64 bit on this platform. It's not meant to be portable code :)

Zac Howland Over a year ago

@adrianmcmenamin Then you can do the same thing, just change the coefficients for the bit shifts.

KitsuneYMG · Accepted Answer · 2014-02-26 01:53:18Z

0

Try something like

unsigned long* buffInL = new unsigned long[2];
char* buffIn=(char*)buffInL;
ifstream inFile(path, ifstream::binary);
inFile.read(buffIn, 8);

Unlike other types, char* is allowed to alias.

answered Feb 26, 2014 at 1:53

KitsuneYMG

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