C ++：通过指针访问struct的成员

Question

I run in to segmentation faults when running this code (no compiler warnings or errors). It occurs when trying to assign "Test" to str->sString

MyClass.cpp
//Constructor
MyClass::MyClass( MyStruct *pDesc )
{   
    pDesc = new MyStruct();

    //This is where I get a segmentation fault  
    pDesc->bar= 0xFF;   

}


MyClass.hpp

class ClGadgetFs
{
    public:
            struct MyStruct{
        int bar;
        };
       MyClass(MyStruct *pDesc = NULL);
};

I thought when calling new I would be aalocating memory for the struct? Like malloc(sizeof(myStruct)) Where am I wrong?

Answer 1

void setStruct(myStruct *str)
{
   str->sString = "Test";
   str->nNo = 4;
}

int main()
{
    myStruct p;
    setStruct(&p);
    return 0;
}

you can do this instead

Edit

int main()
{
   MyStruct *pDesc;
   MyClass myClassInstance( pDesc );
   std::cout<< pDesc->bar << std::endl;
   return 0;
}

and

MyClass::MyClass( MyStruct *pDesc ) 

should be changed to

MyClass::MyClass( MyStruct *& pDesc )

Answer 2

void setStruct(myStruct*& str)

以上可能就是您想要的：更改传递的指针作为输出参数。

Answer 3

#include <string>

struct myStruct
{
   std::string sString;
   int nNo;
};


void setStruct(myStruct **str)
{
   *str = new myStruct();
   (*str)->sString = "Test";
   (*str)->nNo = 4;
}

int main()
{
    myStruct *p;
    setStruct(&p);
}

should be what you want, this is the C-style of passing the pointer; since you're allocating memory for the passed pointer, passing the pointer alone doesn't work, you should pass the pointer's address. Another way would be a reference to the pointer that Joop Eggen's answer points out.

Answer 4

str in the function setStruct is a local variable, whose life time is limited in this function.

So when new returns the address, there is no effect on the actual parameter. It is just the same to

void func(int a){
     a = 4
}

You should use pointer to pointer or reference

void setStruct(myStruct ** str){
    (*str) = new myStruct();
    (*str)->sString = "Test";
    (*str)->nNo = 4;
}

void setStruct(myStruct *& str){
    str = new myStruct();
    str->sString = "Test";
    str->nNo = 4;
}

Answer 5

It's likely that the caller of setStruct is allocating a myStruct on the stack :

myStruct value;

and you are calling setStruct(&value);

That would cause the segmentation fault as you would be attempting to rearrange the stack memory. It's difficult to say anything else without the complete example though.

There's nothing wrong with the code as it stands aside from the fact that the value of the pointer following str = new myStruct(); is not passed back to the caller: the caller would be still referring to the pointer pointing to unallocated memory which would result in undefined behaviour . But that is not causing your crash since given where you say the error happens.

A simple fix would be to change the function prototype to

void setStruct(myStruct*& str)

ie pass the pointer by reference so the caller get's the modified pointer value back.

Answer 6

You should use a reference to your pointer to modify it in your function:

struct myStruct{
    std::string sStrnig;
    int nNo;
};


void setStruct(myStruct* &str){

    str = new myStruct();

    str->sString = "Test";
    str->nNo = 4;
}

main(){
    struct myStruct *str = 0;
    setStruct( str );
}