在typedef-ed结构体上使用sizeof运算符 [英] Using sizeof operator on a typedef-ed struct

问题描述

这可能是太明显了。但是，我找不到具体的答案，虽然许多stackoverflow线程谈论这方面的不同方面。

  typedef struct _tmp {
 unsigned int a; 
 unsigned int b; 
} tmp; 
 
 int main（）
 {
 int c = 10; 
 if（c <= sizeof tmp）{
 printf（less\\\
）; 
} else {
 printf（more\\\
）; 
} 
 return 0; 
} 
  

我将此编译为 -

  g ++ -lstdc ++ a.cpp 
  

得到一个错误 -

 预期的主表达式'）'token 
  pre> 
 
 
我想我缺少一些非常明显和直接的东西。但是似乎无法确定： -  / 
 
 
 
谢谢！ 
解决方案
 
 5.3.3 Sizeof [expr.sizeof] 
 
 
 
 
 
 1） sizeof 运算符产生其操作数的对象表示中的字节数。操作数是
，这是一个未求值的操作数（第5条），或带括号的 type-id 。 
 
在您的情况下，它是一个 type-id ，因此必须括号括起来。在 8.1类型名称[dcl.name] 中描述了一个type-id。
 
 
 
  sizeof tmp 应为 sizeof（tmp）。
 
 
 
如
 
 
 
  if（c <= sizeof tmp）应该 if（c <= sizeof（tmp） code>。
 
 
 
 Yup，非常明显且直接。
 
This might be something too obvious. However, I couldn't find the specific answer though many stackoverflow threads talk about different aspects of this.
typedef struct _tmp {
   unsigned int a;
   unsigned int b;
} tmp;

int main()
{
    int c=10;
    if (c <= sizeof tmp) {
       printf("less\n");
    } else {
       printf("more\n");
    }
    return 0;
}
I compile this prog as - 
g++ -lstdc++ a.cpp
I get an error - 
expected primary-expression before ‘)’ token
I think I am missing something very obvious and straightforward. But can't seem to pinpoint it :-/

Thanks! 
解决方案
5.3.3 Sizeof [expr.sizeof]


  
1) The sizeof operator yields the number of bytes in the object representation of its operand. The operand is
  either an expression, which is an unevaluated operand (Clause 5), or a parenthesized type-id. (emphasis mine)
In your case, it is a type-id so it must be parenthesized. What a type-id is is described in 8.1 Type names [dcl.name].

sizeof tmp should be sizeof (tmp).

As in

if (c <= sizeof tmp) should be if (c <= sizeof (tmp)).

Yup, pretty "obvious and straightforward".

                        
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