How to shift bytes from char array into int

Question

I would like to make a int varibale out of a char array in C.

The char array looks like this:

buffer[0] = 0xcf
buffer[1] = 0x04
buffer[2] = 0x00
buffer[3] = 0x00

The shifting looks like this

x = (buffer[1] << 8 )| (buffer[0] << 0) ;

After that x looks like this:

x = 0xffff04cf

Right now everthing would be fine, if the first two bytes wouldn't be ff.

If I try this line

x = (buffer[3] << 24 )| (buffer[2] << 16)| (buffer[1] << 8)| (buffer[0] << 0) ;

it still looks

x = 0xffff04cf

Even when I try to shift in the zeros before or after I shift in 04cf it looks still the same.

Is this the rigth idea to it or what am I doing wrong?

Pretty sure you meant <<0 and not <<4. When a signed char is converted to signed int the sign is extended, that’s why you get this. You can and the bits or use unsigned values — Sami Kuhmonen
– Sami Kuhmonen, Commented Nov 23, 2017 at 16:50
i wont to get an unsigned int . the buffer comes from recv function of a socket — Gora
– Gora, Commented Nov 23, 2017 at 16:55
Why are you shifting buffer[0] by 8 and buffer[1] by 4? Also this does not result in the value 0xffffcf04, it results in 0xffffcf40. — Christian Gibbons
– Christian Gibbons, Commented Nov 23, 2017 at 17:10
As always, post a minimal reproducible example. Otherwise people need to guess what your actual types are. And as a rule of thumb, 1) use unsigned types for bitwise arithmetic, and 2) use signed types for arithmetics. — vgru
– vgru, Commented Nov 23, 2017 at 17:22

Stephan Lechner · Accepted Answer · 2017-11-23 17:14:36Z

3

The issue is that you declared buffer by means of a signed type, probably (signed) char. When applying operator <<, integral promotions will be performed, and as the value 0xcf in an 8-bit signed type represents a negative value (i.e. -49), it will remain a negative value (yet represented by more bits, i.e. 0xffffffcf). Note that -1 is represented as 0xFFFFFFFF and vice versa.

To overcome this issue, simply define buffer as

unsigned char buffer[4]

And if you weren't allowed to change the data type of buffer, you could write...

unsigned x = ( (unsigned char)buffer[0] << 8 )| ((unsigned char)buffer[1] << 4) ;

edited Nov 23, 2017 at 17:14

answered Nov 23, 2017 at 16:58

Stephan Lechner

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3 Comments

Gora Over a year ago

the buffer is filled with the recv function of winsockets which needs a char*. your suggestion would be fine but the compiler doesn't like it. error C2664: 'recv' : cannot convert parameter 2 from 'unsigned char [512]' to 'char *'

Stephan Lechner Over a year ago

Have you tried the (unsigned char)buffer[0]-thing (without redeclaring buffer as unsigned char buffer[4]?

Gora Over a year ago

it worked with this line : x = (unsigned char) (recvbuf[3] << 24) | (unsigned char)(recvbuf[2] << 16) | ((unsigned char)recvbuf[1] << 8 ) | (unsigned char) (recvbuf[0] << 0) ; thank you :)

PJRobot · Accepted Answer · 2017-11-23 17:36:27Z

0

For tasks like this I like using unions, for example:

union tag_int_chars {
    char buffer[sizeof(int32_t)];
    int32_t value;
} int_chars;

int_chars.value = 0x01234567;

int_chars.buffer[0] = 0xff;

This will automate the memory overlay without the need to shift. Set the value of the int and voila the chars have changed, change a char value and voila the int has changed.

The example will leave the int value = 0x012345ff on a little endian machine.

Another easy way is to use memcpy():

#include <string.h>

char buffer[sizeof(int32_t)];
int32_t value;

memcpy(&value, buffer, sizeof(int32_t));   // chars to int
memcpy(buffer, &value, sizeof(int32_t));   // int to chars

edited Nov 23, 2017 at 17:36

answered Nov 23, 2017 at 16:48

PJRobot

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4 Comments

Sami Kuhmonen Over a year ago

If they really want the bytes in the order they show this won’t work. Also endianness may be an issue. Otherwise it would work

PJRobot Over a year ago

I assumed they wanted to do this through the textual description, but yeah the code does suggest otherwise. Agreed, the user would have to be aware of the endianness for this to work.

Gora Over a year ago

somehow it works, but i don't understand why. Isn't there an easier solution? Converting a 4 Byte char array into an int should be a everyday task and therefore the solution should be somehow "every day" ;)

PJRobot Over a year ago

It works because all fields within a union are aligned to the same memory address, the union sizes to the largest field. In this case the int and the four characters share the same four bytes in memory. This is just one way of doing it, another easy was would be to use memcpy, editing answer to include this.

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