Question

当我尝试在C中实现min函数时，因为C缺少此函数，编译器会给出错误。我已经使用了类型来实现它，就像这样

供快速参考：

>#define min(a,b) \
  ({ typeof (a) _a = (a); \
  typeof (b) _b = (b); \        //Min defined
  _a < _b ? _a : _b; })
  .
  .
  .
  c= min(c+floor(1.0/f), bi.biWidth) ;
  r= min(r+floor(1.0/f),abs(bi.biHeight)) ;

生成错误：

 >make resize
 clang -fsanitize=integer -fsanitize=undefined -ggdb3 -O0 -std=c11 -Wall       
 -Werror -Wextra -Wno-sign-compare -Wshadow    resize.c  -lcrypt -lcs50      
 -lm  -o resize

resize.c:191:18: error: implicit declaration of function 'typeof' is    
invalid in C99 [-Werror,-Wimplicit-function-declaration]
          c= min(c+floor(1.0/f), bi.biWidth) ;
             ^
resize.c:11:6: note: expanded from macro 'min'
({ typeof (a) _a = (a); \
 ^
 resize.c:191:18: error: expected ';' after expression
 resize.c:11:17: note: expanded from macro 'min'
 ({ typeof (a) _a = (a); \
            ^

 resize.c:191:18: error: use of undeclared identifier '_a'

 resize.c:11:17: note: expanded from macro 'min'
 ({ typeof (a) _a = (a); \
            ^

 resize.c:191:18: error: expected ';' after expression

 resize.c:12:18: note: expanded from macro 'min'
  typeof (b) _b = (b); \

                  ^

 resize.c:191:18: error: use of undeclared identifier '_b'

 resize.c:12:18: note: expanded from macro 'min'
  typeof (b) _b = (b); \
             ^

 resize.c:191:18: error: use of undeclared identifier '_a'

 resize.c:13:5: note: expanded from macro 'min'
 _a < _b ? _a : _b; })
 ^

 resize.c:191:18: error: use of undeclared identifier '_b'

 resize.c:13:10: note: expanded from macro 'min'
 _a < _b ? _a : _b; })
     ^

 resize.c:191:18: error: use of undeclared identifier '_b'

 resize.c:13:20: note: expanded from macro 'min'
 _a < _b ? _a : _b; })
               ^

 resize.c:193:13: error: expected ';' after expression
     r= min(r+floor(1.0/f),abs(bi.biHeight)) ;
        ^

 resize.c:11:17: note: expanded from macro 'min'
  ({ typeof (a) _a = (a); \
            ^

 resize.c:193:13: error: use of undeclared identifier '_a'

 resize.c:11:17: note: expanded from macro 'min'
 ({ typeof (a) _a = (a); \
            ^

 resize.c:193:13: error: expected ';' after expression

 resize.c:12:18: note: expanded from macro 'min'
  typeof (b) _b = (b); \
             ^

 resize.c:193:13: error: use of undeclared identifier '_b'

 resize.c:12:18: note: expanded from macro 'min'
  typeof (b) _b = (b); \
             ^

 resize.c:193:13: error: use of undeclared identifier '_a'

 resize.c:13:5: note: expanded from macro 'min'
 _a < _b ? _a : _b; })
 ^

 resize.c:193:13: error: use of undeclared identifier '_b'

 resize.c:13:10: note: expanded from macro 'min'
 _a < _b ? _a : _b; })
     ^

 resize.c:193:13: error: use of undeclared identifier '_b'

 resize.c:13:20: note: expanded from macro 'min'
 _a < _b ? _a : _b; })
               ^

 15 errors generated.

Answer 1

应该是：

#define min(a,b) \
   ({ __typeof__ (a) _a = (a); \
      __typeof__ (b) _b = (b); // Min defined \
      _a < _b ? _a : _b; })

您的评论不应该在\之后但在此之前。你也写下没有下划线的typeof。建议使用它们，给出解释here at SO。

typeof()，__typeof__()和__typeof()是C语言的特定于编译器的扩展，因为标准C不包含此类运算符。

没有typeof的宏：

如果你的编译器不支持typeof“运算符”，你可以定义没有typeof的min宏：

#define min(a,b) (((a) < (b)) ? (a) : (b))

但是，在变量上使用postfix或前缀/减量运算符时会产生副作用，因为会发生双重评估。

内联功能：

使其安全的另一种可能性是内联函数。它将是类型安全的，不会发生双重评估：

inline int min(int a, int b)
{
   if (a < b)
   {
      return a;
   }
   return b;
}

你失去了类型的通用性，但没有别的办法。您也可以考虑使用double，因为整数将被隐式转换。但要记住精确度的松散。

使用typeof编译器扩展在C中实现min函数会产生错误

1 个答案: