Question

我有以下代码：

char switch_list[] = {
    "PINB >> 7", 
    "PIND >> 1", 
    "PINB >> 1", 
    "PIND >> 0}"
};

void values(void){
    uint8_t switch_value = 0;
        if (i == 0){
            switch_value = (PINB >> 7) & 1; 
        }
        if (i == 1){
            switch_value = (PIND >> 1) & 1;
        }
        if (i == 2){
            switch_value = (PINB >> 1) & 1;
        }
        if (i == 3){
            switch_value = (PIND >> 0) & 1;
        }
        SOME OTHER OPERATIONS GO HERE
}

我需要以某种方式将 switch_list 值解释为无符号整数，但我无法对数组进行任何更改（它需要保留为char数组）。 PINB 和其他人已在库中定义了8位值。我想创建一个看起来像这样的for循环：

uint8_t switch_value = 0;
    for (int i = 0, i < sizeof(switch_list)/sizeof(switch_list[0]); i++){
            switch_value = **********[i] & 1; 
         SOME OTHER OPERATIONS GO HERE
        }
}

*********与 switch_list 相同，但不是char类型，而是 uint8_t 。任何人都可以提供任何提示吗？

Answer 1

您可以使用有关数组的知识并创建一个函数，将值从"PINB >> 7"转换为PINB >> 7。我做的假设是：

字符串始终以＆＃34; PIN＆＃34;然后有一个＆＃34; B＆＃34;或者＆＃34; D＆＃34; （可以轻松修改）
然后该字符串将执行操作（目前我只支持＆＃34;＆gt;＆gt;＆＃34;但这也很容易修改）
字符串中的最后一个字符是1个字符编号（同样，可以根据您对字符串的了解进行修改）

使用它，我可以创建一个convert函数

unsigned int convert(char * p);

/* PINB and the others have defined 8 bit value in the libraries
   so I'm making up their values here for convenience */
unsigned int PINB = 1024;
unsigned int PIND = 2048;

int main(){
    // deleted your ending }
    // and changed the type of the array
    char* switch_list[] = {
        "PINB >> 7", 
        "PIND >> 1", 
        "PINB >> 1", 
        "PIND >> 0"
    };

    unsigned int switch_value;
    // , should be ;
    // don't compare signed with unsigned
    for (unsigned int i = 0; i < sizeof(switch_list)/sizeof(switch_list[0]); i++){
        switch_value = convert(switch_list[i]); 
        printf("%u\n", switch_value);
    }

    return 0;
}

// assuming string must be exactly long as "PINB >> 7"
unsigned int convert(char * p){
    if(!p || strlen(p) != strlen("PINB >> 7")){
        printf("error\n");
        return (unsigned)-1;
    }

    unsigned int n;
    // use a string compare or, in your case, since only the 4th char is different:
    if(p[3] == 'B')
        n = PINB;
    if(p[3] == 'D')
        n = PIND;
    // note I'm not handling a case where the 4th letter isn't {'B', 'D'}, according to my assumption (the 1st).

    // use your knowledge about the string inside switch_list
    return n >> (p[strlen(p) - 1] - '0');
}

Answer 2

假设PINx项将评估为PIN_T类型，您可以这样做：

#include <stdlib.h> /* for size_t */
#include <inttypes.h> /* for uint8_t */

/* other include here */


struct switch_s
{
  PIN_T * ppin;
  uint8_t offset;
};

struct switch_s switches[] =
{
  {&PINB, 7},
  {&PIND, 1},
  {&PINB, 1},
  {&PIND, 0},
  /* more here */
};

int main(void)
{ 
  for (size_t i; i < sizeof switches / sizeof *switches; ++i)
  {
    uint8_t switch_value = (*switches[i].ppin >> switches[i].offset) & 1;

    /* Work with switch_value here ... */
  }
}

如何将数组中的字符串转换为无符号整数？

2 个答案: