从（long double *）转换为（const double *） [英] Cast from (long double*) to (const double*)
问题描述
我如何从（long double *）转换为（const double *）
我试过：
const double * Value1 =（const double *）Value2;
编译器不会抱怨，但是当我访问
const double *时实际结果是不正确的。
注意（long double *）是一个指向长双精度数组的指针。
How can I cast from (long double*) to (const double*)
I have tried:
const double* Value1 = (const double*)Value2;
The compiler does not complain but the actual results when I access
the const double* are incorrect.
Note that the (long double*) is a pointer to an array of long doubles.
推荐答案
ferran < FE ***** @ yahoo.com>在消息中写道
news：53 ************************** @ posting.google.c om ...
"ferran" <fe*****@yahoo.com> wrote in message
news:53**************************@posting.google.c om...
如何从（long double *）转换为（const double *）
我试过：
const double * Value1 =（const double *）值2;
你已经回答了自己的问题。
编译器没有抱怨，但是当我访问const时实际结果是double *不正确。
不足为奇。
注意（long double *）是一个指向长双精度数组的指针。
How can I cast from (long double*) to (const double*)
I have tried:
const double* Value1 = (const double*)Value2;
Well you''ve answered your own question.
The compiler does not complain but the actual results when I access
the const double* are incorrect.
Not surprisingly.
Note that the (long double*) is a pointer to an array of long doubles.
我想你要问的问题是如何*将一个长的
双数组转换为一系列双打，投射无益因为你已经发现了
。
分配新的双打数组并复制长双打的唯一方法
一个接一个地到新阵列。类似
double * Value1 = new double [N];
for（int i = 0; i< N; ++ i）
Value1 [i] = Value2 [i];
john
I think the question you meant to ask is how to I *convert* an array of long
doubles to an array of doubles, casting is not helpful as you have already
found out.
The only way to allocate a new array of doubles and copy the long doubles
over to the new array one by one. Something like
double *Value1 = new double[N];
for (int i = 0; i < N; ++i)
Value1[i] = Value2[i];
john
" ferran" < FE ***** @ yahoo.com>在消息中写道
news：53 ************************** @ posting.google.c om ...
"ferran" <fe*****@yahoo.com> wrote in message
news:53**************************@posting.google.c om...
如何从（long double *）转换为（const double *）
我试过：
const double * Value1 =（const double *）值2;
你已经回答了自己的问题。
编译器没有抱怨，但是当我访问const时实际结果是double *不正确。
不足为奇。
注意（long double *）是一个指向长双精度数组的指针。
How can I cast from (long double*) to (const double*)
I have tried:
const double* Value1 = (const double*)Value2;
Well you''ve answered your own question.
The compiler does not complain but the actual results when I access
the const double* are incorrect.
Not surprisingly.
Note that the (long double*) is a pointer to an array of long doubles.
我想你要问的问题是如何*将一个长的
双数组转换为一系列双打，投射无益因为你已经发现了
。
分配新的双打数组并复制长双打的唯一方法
一个接一个地到新阵列。类似
double * Value1 = new double [N];
for（int i = 0; i< N; ++ i）
Value1 [i] = Value2 [i];
john
I think the question you meant to ask is how to I *convert* an array of long
doubles to an array of doubles, casting is not helpful as you have already
found out.
The only way to allocate a new array of doubles and copy the long doubles
over to the new array one by one. Something like
double *Value1 = new double[N];
for (int i = 0; i < N; ++i)
Value1[i] = Value2[i];
john
2004年4月10日星期六16:44:44 +0100 in comp.lang.c ++，John Harrison
< jo ************* @ hotmail.com>写道，
On Sat, 10 Apr 2004 16:44:44 +0100 in comp.lang.c++, "John Harrison"
<jo*************@hotmail.com> wrote,
分配新的双打数组并将长双打的唯一方法是逐个复制到新数组。类似
double * Value1 = new double [N];
for（int i = 0; i< N; ++ i）
Value1 [i] = Value2 [i];
The only way to allocate a new array of doubles and copy the long doubles
over to the new array one by one. Something like
double *Value1 = new double[N];
for (int i = 0; i < N; ++i)
Value1[i] = Value2[i];
肯定有比复制一个更好的东西。
std :: vector< double> Value1（Value2，Value2 + N）;
或最坏的
std :: copy（Value2，Value2 + N，Value1）;
为什么你会为此写一个''for''循环？我不明白。
Surely there is something better than copying one by one.
std::vector<double> Value1( Value2, Value2+N );
Or at the worst
std::copy(Value2, Value2+N, Value1);
Why would you ever write a ''for'' loop for that? I don''t get it.
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