C / C ++：将char []转换为int失败，将unsigned char []转换为int可行，为什么？

Question

I haven't found a question answering this exact behaviour, and somehow I just don't understand what is going on:

I read the contents of a Windows Bitmap File (bmp) into a array and use this array later to extract required information:

char biHeader[40];
// ...
source.read(biHeader,40);
// ...
int biHeight = biHeader[8] | (biHeader[9] << 8) | (biHeader[10] << 16) | (biHeader[11] << 24);

After this, biHeight shows as -112 which is totally wrong because it should be 400 . So, I took a look at a hexdump of the file. The contents read are:

90 01 00 00

Changing the byte order to big endian gives 0x190 which is 400 in decimal, as expected.

If I change above code to:

unsigned char biHeader[40];
// ...
source.read((char*)biHeader,40);
// ...
int biHeight = ... (same as before)

... then I get the expected value. What is going on here?

And: How would you read this data?

Answer 1

As a signed 8-bit two's complement integer, 0x90 is -112 . When that is converted to int for the | , its value is preserved. Since all bits from the seventh on are set if the representation is two's complement, a bitwise or with values shifted left by at least eight bits doesn't change the value anymore.

As an unsigned 8-bit integer, the value of 0x90 is 144, a positive number with no bits beyond the 2^7 bit set. Then, a bitwise or with biHeader[9] << 8 changes the value to the desired 144 + 256 = 400 .

When working with bitwise operators, (almost) always use unsigned types, signed types often lead to unpleasant surprises (and undefined behaviour if the shift result is out of range or a negative integer is shifted left).