MISRA违规规则10.1和枚举

Question

首先,这类似于:如何隐式转换整数类型？但是有一个不同的MISRA警告.

编译器不会生成MISRA错误,但静态分析工具会生成错误.我有一张正在进行工具制造商的票.

鉴于:

#include <stdio.h>
enum Color {RED, VIOLET, BLUE, GREEN, YELLOW, ORANGE};

int main(void)
{
  enum Color my_color;
  my_color = BLUE;
  if (my_color == YELLOW)  // Generates MISRA violation, see below.
  {
     printf("Color is yellow.\n");
  }
  else
  {
     printf("Color is not yellow.\n");
  }
  return 0;
}

静态分析工具正在为if语句生成MISRA违规:

MISRA-2004 Rule 10.1 violation: implicitly changing the signedness of an expression.
Converting "4", with underlying type "char" (8 bits, signed),
to type "unsigned int" (32 bits, unsigned) with different signedness.

编译器是正确的(不识别缺陷)还是静态分析工具？

Answer 1

根据C语言规范,表达式的类型:

typedef enum Colors {RED, VIOLET, BLUE, GREEN, YELLOW, ORANGE} Colors_t;  

是一个signed int.

同样根据语言,枚举项的值是可以包含整个枚举的最小单位.所以在上面的枚举中,BLUE有类型signed char.

当变量与以下项Colors_t比较时,静态分析工具报告MISRA违规BLUE:

Colors_t my_color;
if (my_color == BLUE) // This generates a MISRA violation.

将违规行为与之进行signed int比较signed char.

此外,枚举项可以与其他枚举类型混合而不会出错,因为枚举不是唯一类型:

typedef enum Tree_Species {REDWOOD, CYPRUS, PALM, OAK, JUNIPER, SEGUARO} Tree_Species_t;
Tree_Species_t my_tree;
my_tree = BLUE;