返回值的地址 [英] address of return value

问题描述

如何在功能中获取返回值的地址？


ex。

int function（int argument）

{

int * arg_address;

int * rtn_address;


arg_address =& argument;

rtn_address = ??;

Hi,

How can I get the address of return value in function?

ex.

int function(int argument)
{
int *arg_address;
int *rtn_address;

arg_address = &argument;
rtn_address = ??;

推荐答案

也许，你可以这样做：


int ss（int i）

{

int * k =& i;

int * j =& i;

return（int）j; //这是演员，因为j是一种类型

int *

}

may be,you can do as following:

int ss(int i)
{
int *k = &i;
int *j = &i;
return (int)j; //this is a cast,becase j is a type of
int*
}

也许，你可以如下：

int ss（int i）

{

int * k =& i;

int * j =& i;

return（int）j; //这是一个演员;


}

maybe,you can as following:
int ss(int i)
{
int *k = &i;
int *j = &i;
return (int)j; //this is a cast;

}

ig ********* @ gmail.com 写道：

如何在函数中获取返回值的地址？

Hi,

How can I get the address of return value in function?

返回值甚至不一定具有

地址。对于足够小的类型，您的编译器可以选择将返回值放在寄存器中。


Alan

It is not necessarily the case that the return value even has an
address. For sufficiently small types your compiler may choose to place
the return value in a register.

Alan

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