How use memcpy to initialize array in struct

Question

I have simple struct:

typedef struct{
double par[4];
}struct_type;

I have also initialize function for it where one argument is a 4 elements array. How properly use memcpy to initalize array in struct? Something like this don't work for me:

 struct_type* init_fcn(double array[4]){

 struct _type* retVal;
 retVal->par=malloc(sizeof(double)*4);
 memcpy(retVal->par,&array);

return retVal;
}

I can init values one by one but i thnik memcpy will be better and faster. Do You have any ideas how to proper do it?

Oh I see, memcpy also requires the size of the array, which is the same as the size given to malloc — Paul Stelian
– Paul Stelian, Commented Feb 12, 2017 at 18:07
sizeof(*double) is nonsense and will result in a compiler error. This isn't a minimal reproducible example — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Feb 12, 2017 at 18:07
More like memcpy(retVal->par, array, sizeof(double)*4). Also in malloc you use sizeof(double) (size of one element) — Paul Stelian
– Paul Stelian, Commented Feb 12, 2017 at 18:08
@PaulStelian - Look closely at the bit I quoted. Where is the asterisk (*) located? — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Feb 12, 2017 at 18:12
@PaulStelian - The edit history shows only the nonsense I quoted. But It was all edited away by now, so this discussion is moot. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Feb 14, 2017 at 11:36

Jonathan Leffler · Accepted Answer · 2017-02-12 18:32:52Z

4

If you want to return a pointer to a new object of type struct_type, then you should create exactly such an object, i.e. use malloc(sizeof(struct_type)) instead of allocating space for any members directly. So your code could look as follows:

struct_type* init_fcn(double array[4]){

    struct_type* retVal;
    retVal = malloc(sizeof(struct_type));
    if (retVal) {
        memcpy(retVal->par,array,sizeof(retVal->par));
    }

    return retVal;
}

edited Feb 12, 2017 at 18:32

Jonathan Leffler

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answered Feb 12, 2017 at 18:12

Stephan Lechner

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4 Comments

Stephan Lechner Over a year ago

@PSkocik: noticed right now and corrected it; thanks.

Arkadiusz Kuleta Over a year ago

Thanks! It work for me. So is 'r->par' first value in par array and '&r-par' is adress of first element?

Petr Skocik Over a year ago

@ArkadiuszKuleta No. The first is the array, which in the context of a function call decays to the address of the first element, and the second is the address of the array, which is numerically the same. In this context, you can use r->par, &r->par, &r->par[0], and it doesn't matter which one you use.

Jonathan Leffler Over a year ago

@PSkocik: More accurately, r->par and &r->par[0] are the same type (double *) and value; &r->par is of type double (*)[4], though its value as a void * is the same as r->par etc.

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