通过按值 [英] pass-by-value
问题描述
我需要你的帮助，因为我不太了解这一点：
C参数中的
按值传递。函数参数得到一个副本
的参数值。
但是如果我传递一个指针究竟发生了什么？还有一份副本通过
？
$ C $ b在C ++中也有一个传递参考...在这种情况下
参数可以被认为是参数的一个
别名...
但是当我们传递一个指针时现在在C中我们可以想到它作为一个
传递地址机制来操纵
变量？
谢谢
Hi, I need your help because I don''t understand very well this:
in C arguments are passed by-value. The function parameters get a copy
of the argument values.
But if I pass a pointer what really is happening? also a copy is passed
?
in C++ there is a pass-by-reference too... and in that case the
paramter can be considered as an
alias of the argument...
but now in C when we pass a pointer can we think of it as a
pass-by-address mechanism to manipulate
variables?
Thanks
推荐答案
xdevel说：
xdevel said:
我需要你的帮助因为我不太了解这个：
C参数中的
按值传递。
Hi, I need your help because I don''t understand very well this:
in C arguments are passed by-value.
正确。
Correct.
函数参数获取参数值的副本。
The function parameters get a copy of the argument values.
更准确地说，参数是*表达式*。参数表达式是
被评估，并且这些值存储在函数的参数中。
More precisely, arguments are *expressions*. The argument expressions are
evaluated, and those values are stored in the parameters to the function.
但是如果我传递指针真的是什么正在发生？
But if I pass a pointer what really is happening?
指针被计算，指针'的值存储在函数的
参数中，与任何参数一样其他论证表达。
-
Richard Heathfield
Usenet是一个奇怪的地方 - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上面的域名（但显然放弃了www）
The pointer is evaluated, and that pointer''s value is stored in the
parameter to the function, just the same as any other argument expression.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
xdevel写道：
xdevel wrote:
我需要你的帮助，因为我不太了解这一点：
in C arguments按值传递。函数参数获得参数值的副本
。
Hi, I need your help because I don''t understand very well this:
in C arguments are passed by-value. The function parameters get a copy
of the argument values.
这是正确的。
That''s correct.
但是如果我传递指针真的是什么正在发生？也是副本通过
？
But if I pass a pointer what really is happening? also a copy is passed
?
传递*指针*的副本。它指向的是*不*
复制。
A copy of the *pointer* is passed. The thing it''s pointing to is *not*
copied.
在C ++中有一个传递-inference也是......在这种情况下，
参数可以被认为是参数的一个
别名...
in C++ there is a pass-by-reference too... and in that case the
paramter can be considered as an
alias of the argument...
虽然OT在这里，C ++引用实际上只是一种不同的方式来引用
到指针。隐藏&的只是语法糖。 （在
电话中）和*（） （在函数内部）来自你的运算符。
While OT here, C++ references are really just a different way to refer
to pointers. It''s just syntactic sugar that hides the "&" (at the
call) and "*()" (inside the function) operators from you.
但是当我们传递指针时现在在C中我们可以把它想象成一个
传递地址机制来操纵
变量？
but now in C when we pass a pointer can we think of it as a
pass-by-address mechanism to manipulate
variables?
如果我正确地解释你的问题，是的。传递的指针
将指向原始对象，原始对象可能是通过该指针访问的
。
If I''m interpreting your question correctly, yes. The passed pointer
will point to the original object, and the original object may be
accessed via that pointer.
< br>
ro *********** @ yahoo。 com ha scritto：
ro***********@yahoo.com ha scritto:
虽然OT在这里，C ++引用实际上只是一种不同的方式来引用
到指针。隐藏&的只是语法糖。 （在
电话中）和*（） （在函数内部）来自你的运营商。
While OT here, C++ references are really just a different way to refer
to pointers. It''s just syntactic sugar that hides the "&" (at the
call) and "*()" (inside the function) operators from you.
ok但是如果我用引用写一个交换函数我可以这样做：
void swap（int& a，int& b）
{
int temp = 0;
int& tmp = temp;
tmp = a;
a = b;
b = tmp;
}
这里我可以交换对象直接
而不是指针我可以这样做：
void swapP（int * a，int * b）
{
int tmp;
tmp = * a;
* a = * b;
* b = tmp;
}
我不能直接交换指针......
所以我的问题这就是为什么如果通过引用可以考虑作为
传递指针
代码是不同的？
什么类型的指针是参考？
ok but if I write a swap function with reference I can do this:
void swap(int &a, int &b)
{
int temp = 0;
int &tmp = temp;
tmp = a;
a = b;
b = tmp;
}
here I can swap the objects directly
and instead with pointer I can do this:
void swapP(int *a, int *b)
{
int tmp;
tmp = *a;
*a = *b;
*b = tmp;
}
I can''t swap directly the pointers...
and so my question is why if pass-by-reference can be considerd as a
pass-by-pointer
is the code different?
what type of pointer is a reference?
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