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How to make a shallow copy of entire structure?

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I want to copy an identical struct into another and later on use it as a comparance to the first one. The thing is that my compiler gives me a warning when Im doing like this! Should I do it in another way or am I doing this wrong:

In header File:

extern struct RTCclk { uint8_t second; uint8_t minute; uint8_t hour; uint8_t mday; uint8_t month; uint8_t year; } RTCclk; 

In C file:

struct RTCclk RTCclk; struct RTCclk RTCclkBuffert;  void FunctionDO(void) {    ... // Some Code    /* Copy first struct values into the second one */    memcpy(&RTCclk, &RTCclkBuffert, sizeof RTCclk); } 

Kind Regards!

Answers

For simple structures you can either use memcpy like you do, or just assign from one to the other:

RTCclk = RTCclkBuffert;

The compiler will create code to copy the structure for you.



An important note about the copying: It's a shallow copy, just like with memcpy. That means if you have e.g. a structure containing pointers, it's only the actual pointers that will be copied and not what they point to, so after the copy you will have two pointers pointing to the same memory.

Your code is correct. You can also assign one directly to the other (see Joachim Pileborg's answer).

When you later come to compare the two structs, you need to be careful to compare the structs the long way, one member at a time, instead of using memcmp; see How do you compare structs for equality in C?

copy structure in c you just need to assign the values as follow:

struct RTCclk RTCclk1;
struct RTCclk RTCclkBuffert;

RTCclk1.second=3;
RTCclk1.minute=4;
RTCclk1.hour=5;

RTCclkBuffert=RTCclk1;

now RTCclkBuffert.hour will have value 5,

RTCclkBuffert.minute will have value 4

RTCclkBuffert.second will have value 3

Your memcpy code is correct.

My guess is you are lacking an include of string.h. So the compiler assumes a wrong prototype of memcpy and thus the warning.

Anyway, you should just assign the structs for the sake of simplicity (as Joachim Pileborg pointed out).

Also a good example.....

struct point{int x,y;};
typedef struct point point_t;
typedef struct
{
    struct point ne,se,sw,nw;
}rect_t;
rect_t temp;

int main()
{
//rotate
    RotateRect(&temp);
    return 0;
}

void RotateRect(rect_t *givenRect)
{
    point_t temp_point;
    /*Copy struct data from struct to struct within a struct*/
    temp_point = givenRect->sw;
    givenRect->sw = givenRect->se;
    givenRect->se = givenRect->ne;
    givenRect->ne = givenRect->nw;
    givenRect->nw = temp_point;
}

memcpy expects the first two arguments to be void*.

Try: memcpy( (void*)&RTCclk, (void*)&RTCclkBuffert, sizeof(RTCclk) );

P.S. although not necessary, convention dictates the brackets for the sizeof operator. You can get away with a lot in C that leaves code impossible to maintain, so following convention is the mark of a good (employable) C programmer.

I think you should cast the pointers to (void *) to get rid of the warnings.

memcpy((void *)&RTCclk, (void *)&RTCclkBuffert, sizeof RTCclk);

Also you have use sizeof without brackets, you can use this with variables but if RTCclk was defined as an array, sizeof of will return full size of the array. If you use use sizeof with type you should use with brackets.

sizeof(struct RTCclk)

What warning do you have ?

Your variable RTCclk have the same name as the struct, thus putting the compiler to an ambiguous evaluation of &RTCclk... Your memcpy is correct but you should write sizeof(RTCclk).

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