通过引用传递const指针 [英] passing const pointer by reference

问题描述

我很困惑为什么后面的代码不能编译

  int foo（const float *& a）{ 
 return 0; 
} 
 int main（）{
 float * a; 
 foo（a）; 
 
 return 0; 
} 
  

编译器给出错误：

$ b b

错误：类型为'float *'的类型为'const float *&'的引用初始化无效

但是当我尝试在foo中不通过引用传递，它是编译良好。

我认为它应该显示相同的行为，不管我是否通过引用。

>

解决方案

因为它不是类型安全的。请考虑：

  const float f = 2.0; 
 int foo（const float *& a）{
 a =& f; 
 return 0; 
} 
int main（）{
 float * a; 
 foo（a）; 
 * a = 7.0; 
 
 return 0; 
} 
  

任何非 - const 引用或指针必须在指向类型中是不变的，因为非< - code> const 指针或引用支持读取（协变操作）以及写入（逆变换操作）。

const 必须首先从最大间接级别添加。这将工作：

  int foo（float * const& a）{
 return 0; 
} 
 int main（）{
 float * a; 
 foo（a）; 
 
 return 0; 
} 
  

I am confused that why following code is not able to compile

int foo(const float* &a) {
    return 0;
}
int main() {
    float* a;
    foo(a);

    return 0;
}

Compiler give error as:

error: invalid initialization of reference of type 'const float*&' from expression of type 'float*'

but when I try to pass without by reference in foo, it is compiling fine.

I think it should show same behavior whether I pass by reference or not.

Thanks,

解决方案

Because it isn't type-safe. Consider:

const float f = 2.0;
int foo(const float* &a) {
    a = &f;
    return 0;
}
int main() {
    float* a;
    foo(a);
    *a = 7.0;

    return 0;
}

Any non-const reference or pointer must necessarily be invariant in the pointed-to type, because a non-const pointer or reference supports reading (a covariant operation) and also writing (a contravariant operation).

const must be added from the greatest indirection level first. This would work:

int foo(float* const &a) {
    return 0;
}
int main() {
    float* a;
    foo(a);

    return 0;
}

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