抓取8字节字的高4字节

Question

I am multiplying 0x1d400 * 0xE070381D .

When I do this on my calculator the result is 0x00019A4D26950400

When I tried implementing this in cpp here's what i have.

long long d;

d = 3765450781 * 1d400;

The result this code gives is that d = 0x26950400 . This is only the bottom 4 bytes, what happened to everything else?

I am trying to isolate the upper 4 bytes 0x00019A4D and save them into another variable. How can this be done?

If I could get the multiplication to display all 8 bytes what I was thinking of doing to isolate the upper 4 bytes was:

d = d & 0xFF00; //0xFF00 == (binary) 1111111100000000

d = d>>8;

Will this work?

Answer 1

在数字后添加LL （例如3765450781LL ），否则它们将被计算为int s，其余部分在分配给d之前被切断。

Answer 2

You need to use LL after your constant for a long long , as indicated in another answer by MByD.

In addition, your data type should be unsigned long long . Oherwise when you right shift, you may get the leftmost bit repeated due to sign-extend. (That is machine-dependent, but most machines sign extend negative numbers when right shifting.)

You do not need to mask off the upper 4 bytes before right shifting, because you are going to throw away the lower 4 bytes any way when you do the right shift.

Finally, note that the argument to >> is the number of bits to shift, not bytes. Therefore, you want

d = d >> 32;

Which can also be written as

d >>= 32;

Answer 3

As the others pointed out, you must suffix your 64-bit numeric literals with LL .

To print your long long variable in hex, use the format specifier "%016llX" :

long long d;
d = 3765450781LL * 0x1d400LL;
printf("%016llX\n", d);

outputs 00019A4D26950400 .

To get the upper and lower 32 bits (4 bytes) of the variable d , you can do:

unsigned int upper;
unsigned int lower;

upper = d >> 32;
lower = d & 0x00000000FFFFFFFF;

printf("upper: %08X lower: %08X\n", upper, lower);