使用 typeof 编译器扩展在 C 中实现 min 函数会出错

Question

当我试图在 C 中实现 min 函数时，因为 C 缺少此函数，编译器给出了一个错误。我使用 type of 来实现它就像这样

MIN and MAX in C

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)) ;

产生错误：

 >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; })

您的评论应该而不是在\之后，而是在它之前。你也写 typeof 没有下划线。推荐使用它们，给出解释here at SO.

typeof(), __typeof__() and __typeof() are compiler-specific extensions to the C language, because standard C does not include such an operator.

没有 typeof 的宏:

如果你的编译器不支持typeof"operator"，你可以定义没有typeof的min宏：

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

但是，在对变量使用后缀或前缀 in/decrement 运算符时，会产生副作用，因为会发生双重计算。

内联函数：

另一种使其安全的可能性是内联函数。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;
}

你失去了类型通用性，但别无他法。您还可以考虑使用 double，因为整数将被隐式转换。但请记住精度的松散。

使用 typeof 编译器扩展在 C 中实现 min 函数会出错

Implementing min function in C by using typeof compiler extension gives error

c

gcc

max

min

clang