问题描述
/ *开始端口到c ++ * /
＃ifdef __cplusplus
externC" ; {
＃endif
int * get_an_int（void）;
int * pass_pointer_triv（int *）;
＃ifdef __cplusplus
}
＃endif
/ *结束端口到c ++ * /
#include< stdio.h>
#include< stdlib.h>
int main（void）
{
int * rt，* qt;
qt = get_an_int（）;
rt = pass_pointer_triv（qt ）;
printf（t is％d \ n，* rt）;
printf（" tja \ n"）;
返回0;
}
int * get_an_int（无效）
{
int t;
int * pt;
pt = malloc（sizeof * pt）;
t = 41;
pt [0] = t;
返回pt;
}
pass_pointer_triv（int * a）
{
int t;
t = * a;
printf（" t is％d \ n"， t）;
返回a;
}
/ *结束来源* /
我相信这个来源是一个愚蠢的错误，远离kosher c ++。
我的编译器告诉我，微不足道的指针传递在
间接的水平上有所不同，但在我看来，这一切都是看起来一样。我相信我的
编译器是正确的。感谢任何提示。 furunculus
/* begin port to c++ */
# ifdef __cplusplus
extern "C" {
# endif
int * get_an_int(void);
int * pass_pointer_triv(int *);
# ifdef __cplusplus
}
# endif
/* end port to c++ */
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int *rt, *qt;
qt = get_an_int();
rt = pass_pointer_triv(qt);
printf ("t is %d\n", *rt);
printf ("tja\n");
return 0;
}
int * get_an_int(void)
{
int t;
int * pt;
pt = malloc (sizeof*pt);
t = 41;
pt[0] = t;
return pt;
}
pass_pointer_triv(int *a)
{
int t;
t = *a;
printf("t is %d\n", t);
return a;
}
/* end source */
I believe that this source is one silly error away from being kosher c++.
My compiler tells me that the trivial pointer pass varies in level of
indirection, but to my eye, it all looks the same. I believe that my
compiler is correct. Grateful for any hints. furunculus
推荐答案
我的编译器抱怨这个。
错误C2440：''=''：无法从''void *''转换为''int *''
无论是强制转换为int *还是使用new（首选）。
My compiler complains about this.
error C2440: ''='' : cannot convert from ''void *'' to ''int *''
Either typecast to int * or use new (preferred) instead.
请解释。
Please explain.
您忘记了返回类型。将此更改为
int * pass_pointer_triv（int * a）
并为它编译，除了malloc错误。
You forgot the return type. Change this to
int* pass_pointer_triv(int *a)
and it compiles for me, except for the malloc error.
我认为那会'让她去谢谢你假装我的问题
今晚是你的问题。 ciao，f
I think that''s gonna get her going. Thanks for pretending like my problems
were your problems tonight. ciao, f
请解释。
Please explain.
malloc返回void *（void指针）。如果你在c ++中使用它，你需要
使用reinterpret_cast或c-style
强制转换为正确的类型。这些工作中的任何一个，reinterpert_cast首选。
pt = reinterpret_cast< int *>（malloc（sizeof * pt））;
pt =（int *）malloc（sizeof * pt）;
但是首选使用new。
int * pt = new int;
那么只需使用delete来摆脱内存而不是免费（你使用带有malloc的
，但你的代码中缺少你的内存泄漏）
malloc returns a void* (void pointer). If you use it in c++ you need to
typecast it to the proper type either with reinterpret_cast or a c-style
cast. Either of these work, the reinterpert_cast prefered.
pt = reinterpret_cast<int*>( malloc( sizeof *pt ) );
pt = (int*) malloc( sizeof *pt );
but prefered is using new.
int *pt = new int;
Then just use delete to get rid of the memory instead of free (which you use
with malloc, but you are missing in your code so you have a memory leak)
您忘记了返回类型。将此更改为
int * pass_pointer_triv（int * a）
并为我编译，除了malloc错误。
You forgot the return type. Change this to
int* pass_pointer_triv(int *a)
and it compiles for me, except for the malloc error.
我认为这会让她继续前进。感谢像我这样的问题假装你今晚的问题。 ciao，f
I think that''s gonna get her going. Thanks for pretending like my
problems were your problems tonight. ciao, f
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