问题描述
在C ++中我有一个函数,只需要只读访问数组,但错误地声明为接收一个非const指针: size_t countZeroes(int * array,size_t count)
{
size_t result = 0;
for(size_t i = 0; i if(array [i] == 0){
++ result;
}
}
返回结果;
}
我需要为一个const数组调用它:
static const int Array [] = {10,20,0,2};
countZeroes(const_cast< int *>(Array),sizeof(Array)/ sizeof(Array [0]))
这将是未定义的行为吗?如果是这样,当程序运行到UB时 - 当执行const_cast并调用functon或访问数组时?
是的,它是允许的(如果危险!)。它实际上是写入到一个引发未定义行为的 const
对象,而不是转换本身(7.1.5.1/4 [dcl.type.cv])。
作为5.2.11 / 7 [expr.const.cast]中的标准注释,根据对象的类型,尝试通过指针写入,该指针是抛出 const
可能会产生未定义的行为。
In C++ I have a function that only requires read-only access to an array but is mistakenly declared as receiving a non-const pointer:
size_t countZeroes( int* array, size_t count )
{
size_t result = 0;
for( size_t i = 0; i < count; i++ ) {
if( array[i] == 0 ) {
++result;
}
}
return result;
}
and I need to call it for a const array:
static const int Array[] = { 10, 20, 0, 2};
countZeroes( const_cast<int*>( Array ), sizeof( Array ) / sizeof( Array[0] ) );
will this be undefined behaviour? If so - when will the program run into UB - when doing the const_cast and calling the functon or when accessing the array?
Yes, it is allowed (if dangerous!). It's the actual write to a const
object that incurs undefined behaviour, not the cast itself (7.1.5.1/4 [dcl.type.cv]).
As the standard notes in 5.2.11/7 [expr.const.cast], depending on the type of the object an attempt to write through a pointer that is the result of casting away const
may produce undefined behaviour.
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