[英]accessing structure member using pointers
I am a newbie in C. I am trying to create a typedef
struct outside of main and then create a pointer of typedef
. 我是C语言的新手。我试图在main外部创建一个typedef
结构,然后创建一个typedef
指针。 Then pass this pointer into another function. 然后将此指针传递给另一个函数。 However I am getting error. 但是我得到了错误。 It is driving me crazy .Thank you very much in advance.. 这让我发疯了。非常感谢您。
typedef struct rem_info
{
char ufrag[80];
char pwd[80];
unsigned comp_cnt;
pj_sockaddr def_addr[PJ_ICE_MAX_COMP];
unsigned cand_cnt;
pj_ice_sess_cand cand[PJ_ICE_ST_MAX_CAND];
} rem_info;
void reset_rem_info(rem_info *prem)
{
pj_bzero(prem, sizeof(rem_info));
}
int main()
{
rem_info *prem;
reset_rem_info(&prem);
return 0;
}
Error: 错误:
*WARNING**:ex7.c:51:1: warning: passing argument 1 of ‘reset_rem_info’ from incompatible pointer type [enabled by default]
reset_rem_info(&prem);
^
ex7.c:41:6: note: expected ‘struct rem_info *’ but argument is of type ‘struct rem_info **’
void reset_rem_info(rem_info *prem)
Looking at your main function: 查看您的主要功能:
int main()
{
rem_info *prem;
reset_rem_info(&prem);
return 0;
}
You are creating a pointer to rem_info and passing its address to reset_rem_info. 您正在创建一个指向rem_info的指针,并将其地址传递给reset_rem_info。 That means you are passing a pointer to a pointer to a rem_info. 这意味着您将传递一个指向rem_info的指针。 To make it typecheck, you could pass the pointer directly without taking its address. 要进行类型检查,可以直接传递指针而无需获取其地址。
int main()
{
rem_info *prem;
reset_rem_info(prem);
return 0;
}
But that will probably give you a bug. 但这可能会给您带来一个错误。 You are now dealing with an uninitialized pointer to rem_info. 您现在正在处理指向rem_info的未初始化的指针。 What you probably want is to create an actual rem_info and pass its address (a pointer to rem_info) to the function. 您可能想要的是创建一个实际的rem_info并将其地址(指向rem_info的指针)传递给该函数。
int main()
{
rem_info prem;
reset_rem_info(&prem);
return 0;
}
void reset_rem_info(rem_info *prem)
Here the function argument expects a pointer of type rem_info and what you are passing is the address of the pointer so there is a type mismatch and hence the warning. 在这里,函数参数需要一个类型为rem_info的指针,而您要传递的是指针的地址,因此存在类型不匹配的情况,因此出现警告。
You can have 你可以有
void reset_rem_info(rem_info **prem)
Make sure you initialize the pointer and pass the address of pointer prem
to pointer to pointer in your function argument. 确保初始化指针,并将指针prem
的地址传递给函数参数中指向指针的指针。 Like shown below 如下图所示
int main()
{
rem_info *prem = malloc(sizeof(rem_info));
reset_rem_info(&rem_info);
}
or while calling the function have 或在调用函数时
int main()
{
rem_info prem;
reset_rem_info(&prem);
}
So that your function prototype stays unchanged. 这样您的函数原型便保持不变。
Thanks guys for the quick replies..Gabriel response gave me a good insight. 谢谢大家的快速回复。.Gabriel的回复给了我很好的见解。 However I stuck to Raphael Santos response. 但是我坚持拉斐尔·桑托斯的回应。 ...However, if Gabriel could kindly elaborate the point a bit more please...ok here is the fixed code ...但是,如果加百列(Gabriel)可以再详细说明一点...好吧,这里是固定代码
typedef struct rem_info
{
char ufrag[80];
char pwd[80];
unsigned comp_cnt;
pj_sockaddr def_addr[PJ_ICE_MAX_COMP];
unsigned cand_cnt;
pj_ice_sess_cand cand[PJ_ICE_ST_MAX_CAND];
}rem_info;
void reset_rem_info(rem_info *prem)
{
pj_bzero(prem, sizeof(rem_info));
}
int main()
{
rem_info prem;
reset_rem_info(&prem);
return 0;
}
The change got rid of the warnings and the segmentation dump... 更改摆脱了警告和细分转储...
THANKS A LOT GUYSSS 感谢很多人
void reset_rem_info(rem_info **prem)
{
pj_bzero(*prem, sizeof(rem_info));
}
int main()
{
rem_info *prem;
reset_rem_info(&prem);
return 0;
}
if you want to malloc prem in reset_rem_info you forgot a * for void reset_rem_info(rem_info *prem) then dereference it for bzero else don t write the & in reset_rem_info(&prem);
如果您想在reset_rem_info中分配prem,您会忘记*为void reset_rem_info(rem_info * prem),然后将其取消引用为bzero,否则不要在reset_rem_info(&prem);
写入& reset_rem_info(&prem);
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