问题
Here is the code:
#define u8 char
#define u32 unsigned int
typedef struct {
//decoded instruction fields
u8 cond; // Condition (f.ex. 1110 for always true)
u8 instruction_code; // Is a constant since we only use branch
u32 offset; // Offset from current PC
} dcdinst;
u8 mem[1024];
mem[0x0] = 0b11101010;
u8* instruction_addr = &mem[pc];
if (instruction_addr == NULL) {
return false;
}
unsigned int first_part = instruction_addr[0];
// Here is the code that presents a problem:
// I try to get the upper part of the first byte
inst.cond = first_part >> 4;
first_part
is the following byte: 11101010.
inst.cond
becomes 11111110, but I need it to be 00001110.
So, my actual problem is that I want to get the first 4 bits of my instruction which starts at the address instruction_addr
. I tried to do so by using the right shift operator >>
but the problem is that instead of prepending 0s to the left of the byte, it prepends 1s.
I found on stackoverflow that I first had to cast the value to a unsigned one, and that's what I did by using the variable first_part
, but I still have the same problem. I don't understand why the shift seems to "see" my variable as a negative one while its type is specifically "unsigned".
Does anybody have an idea?
回答1:
Your u8
type is using char
without specifying signedness, which means it has undefined signedness. It's likely your compiler is using a signed char
by default. Thus, you're subject to sign-extension during operations and during promotion.
Change:
#define u8 char
#define u32 unsigned int
to:
typedef unsigned char u8;
typedef unsigned int u32;
(or use stdint.h
types properly), and your storage should be actually unsigned.
Using typedef
s also means the compiler is involved with this aliasing, it's not just a preprocessor text replacement, eliminating a class of subtle errors.
来源:https://stackoverflow.com/questions/43239906/shift-operator-in-c-prepends-ones-instead-of-zeros