c - 我想在 MPLAB XC8 上获得一点 char 但我不能？

Question

我的功能是：

extern volatile unsigned char Temp       @ 0x036;
extern volatile __bit W       @ (((unsigned) &Temp)*8) + 4;

void get_bit(volatile unsigned char *reg, unsigned num) {
    W = (*reg & (1 << num));
}

主要功能是：

int main() {
    volatile unsigned char ch = 0b00001000;
    get_bit(&ch, 4);
}

当我编译这个块代码时，我得到一个错误（错误：表达式语法）。

我能做些什么来解决这个问题？

Answer 1

试试这个代码：

#include <stdio.h>

unsigned char get_bit(unsigned char reg, unsigned num) 
{
    return (reg & (1 << num));
}

unsigned char get_bit2(unsigned char reg, unsigned num) 
{
    return (reg & (1 << num))?1:0;
}

int main() 
{
    volatile unsigned char ch = 0b00001000;

    ch |= (1<<4);   // To set bit 4
    printf("%d\n",get_bit(ch, 4)); // If you try on a PC
    printf("%d\n",get_bit2(ch, 4)); // If you try on a PC

    ch &= (~(1<<4));   // To reset bit 4
    printf("%d\n",get_bit(ch, 4)); // If you try on a PC
    printf("%d\n",get_bit2(ch, 4)); // If you try on a PC


    return 0;
}