屏蔽数组成员的位

Question

你能告诉我我的问题在哪里吗？我正在尝试测试数组的值。

每个值都有 16 位（标志），我需要检查标志是否已设置。

我的当前输出以及预期输出如下所示。

 #include <stdio.h>

    unsigned short value[1];
    unsigned int bitCheck(unsigned int mask, int pin);
    unsigned short mask;

int main(void){
    value[0]=0;
    value[1]=4095;


    int pin0 = 0;
    int pin1 = 1;

    unsigned int bit0= bitCheck( mask, pin0);
    unsigned int bit1= bitCheck( mask, pin1);


  for (int i =0;i<=1; i++)
{
    mask=value[i];
   printf("Mask = %d ==>>\n", mask);

   if ( bit0 == 1 ){
      printf("Pin %d is Set\n", pin0);
   }else{
      printf("Pin %d is not Set\n", pin0);
   }

    if ( bit1 == 1 ){
      printf("Pin %d is Set\n", pin1);
   }else{
      printf("Pin %d is not Set\n", pin1);
   }

   printf("\n"); 
}

}

unsigned int bitCheck(unsigned int mask, int bit){
   if ( (mask >> bit ) & 1){
      return 1;
   }else{
      return 0;
   }
}

我的输出是：

Mask = 0 ==>>
Pin 0 is not Set
Pin 1 is not Set

Mask = 4095 ==>>
Pin 0 is not Set
Pin 1 is not Set

而且必须是：

Mask = 0 ==>>
Pin 0 is not Set
Pin 1 is not Set

Mask = 4095 ==>>
Pin 0 is Set
Pin 1 is Set

Answer 1

这段代码不必要地复杂，并且通过它的复杂性，隐藏了一个明显的逻辑错误。在给 mask.

赋值之前，您正在调用 bitCheck(mask, ...)
改变这个：

unsigned int bit0= bitCheck( mask, pin0); unsigned int bit1= bitCheck( mask, pin1); for (int i =0;i<=1; i++) { mask=value[i]; printf("Mask = %d ==>>\n", mask);

对此：

for (int i =0;i<=1; i++) { mask=value[i]; unsigned int bit0= bitCheck( mask, pin0); unsigned int bit1= bitCheck( mask, pin1); printf("Mask = %d ==>>\n", mask);

就是说，这里有一个理智的方法来完成这一切：

#include <stdio.h> #define BIT(n, v) (!!((v) & (1U << (n)))) int main(void) { unsigned short values[] = {0, 4095}; for (int i = 0; i < 2; ++i) { printf("value: %hu\n", values[i]); for (int bit = 0; bit < 16; ++bit) { printf( BIT(bit, values[i]) ? "bit %d is set\n" : "bit %d is not set\n", bit); } } return 0; }

您的方法被认为不合理的原因之一，请阅读 DRY principle。编写 16 次相同的代码是违反此原则的一个简单示例 -> 你需要一个循环。

屏蔽数组成员的位

Masks bits of array members

c

arrays

embedded

flags

bit-manipulation