c ++如何解释二进制数

Question

I have a general question about binary numbers in c++. I am reading in a binary file of 32-bit numbers, and then writing these numbers to a text file. My question is, when I do

long int temp;
temp = ( fileBuf[N * 4 * i + 4 * j + 0] << 24 |
         fileBuf[N * 4 * i + 4 * j + 1] << 16 |
         fileBuf[N * 4 * i + 4 * j + 2] << 8  | 
         fileBuf[N * 4 * i + 4 * j + 3] << 0  );
myfile1 << temp << "\t";

does c++ understand that I want it to reinterpret the binary as a decimal number?

Answer 1

All numbers inside of a C++ int variable are binary. The conversion to or from decimal occurs as the number is read or written, or as the compiler converts it from a constant into code.

Answer 2

If I interpret this question as "Will this do what I want it to?" then the answer is yes , as long as what you want it to do is read a 32 bit, big endian integer out of a buffer (presumably loaded from a file) at offset 4 * N * i + 4 * j .

Assuming of course that fileBuf is declared as a character or unsigned character type. It would behave differently if it were, for example, an array of shorts. You'd end up with a mutilated representation of a 64 bit quantity.

Answer 3

实际上，在最后一行打印之前，该数字不会转换为十进制数。