Copying to a dynamically-allocated array in C

Question

I'm trying to copy to a dynamically-allocated array in C (The source array may or may not be dynamically allocated. The first element of the array is copied correctly, but the rest is not. Can anyone tell what I'm doing wrong?

Here's where I'm calling the method:

vec3 tempVel = {
    (float)rand()/(float)RAND_MAX * (2.0f * INIT_VELOCITY_DIMENSION_MAGNITUDE) - INIT_VELOCITY_DIMENSION_MAGNITUDE,
    (float)rand()/(float)RAND_MAX * (2.0f * INIT_VELOCITY_DIMENSION_MAGNITUDE) - INIT_VELOCITY_DIMENSION_MAGNITUDE,
    (float)rand()/(float)RAND_MAX * (2.0f * INIT_VELOCITY_DIMENSION_MAGNITUDE) - INIT_VELOCITY_DIMENSION_MAGNITUDE
};
thisBoid->velocity = copyArrays2(&tempVel, 3);

And the method:

vec3* copyArrays2(vec3 *src, int len){

    vec3 *tempArray = (vec3*)malloc(len*sizeof(vec3));
    for(int i=0; i<len; i++){
        *tempArray[i] = *src[i];
    }

    return tempArray;
}

Printing the values of 'tempVel' before copying and 'thisBoid->velocity' afterwards gives the following output:

{0.895216, -0.107424, 0.532019}
{0.895216, 10.000000, 0.500000}

EDIT: 10 and .5 are always copied regardless of the contents of tempVel — jtruskow, Jun 01 '15 at 14:33
There's no way this compiles, you're dereferencing a `vec3`... @iharob ..."please" ? :) — Quentin, Jun 01 '15 at 14:36
tempVel? It has 3 floats that are separated by commas. A vec3 object inherently has 3 elements — jtruskow, Jun 01 '15 at 14:37
[Do not cast the return value of `malloc()`](http://stackoverflow.com/a/605858/1983495), and use [`memcpy()`](http://man7.org/linux/man-pages/man3/memcpy.3.html) — Iharob Al Asimi, Jun 01 '15 at 14:38
Probably `*tempArray[i] = *src[i];` --> `tempArray[i] = src[i];` — BLUEPIXY, Jun 01 '15 at 14:40
@BLUEPIXY I tried that, I get the error: "array type 'vec3' (aka 'float [3]') is not assignable" — jtruskow, Jun 01 '15 at 14:41
@user3638238 The confusion stems from the fact that we did not know that `vec3` was actually a typedef for `float[3]`... Which now makes sense. — Quentin, Jun 01 '15 at 14:42
if output as `{0.895216, -0.107424, 0.532019} {0.895216, 10.000000, 0.500000}`, `vec3 tempVel[] ={{0.895216, -0.107424, 0.532019}, {0.895216, 10.000000, 0.500000}}; thisBoid->velocity = copyArrays2(tempVel, 2);` — BLUEPIXY, Jun 01 '15 at 14:46
@Quentin that is correct, sorry about that. So how would I go about accessing the ith index of a vec3? — jtruskow, Jun 01 '15 at 14:50
@user3638238 just like Peter Schneider did, use the subscript (`[]`) operator. Or, as hinted by iharob, calculate the total size of your array and use `memcpy`. — Quentin, Jun 01 '15 at 14:52
@user3638238 I had assumed like as `typedef struct vec3 { float v[3]; } vec3;` — BLUEPIXY, Jun 01 '15 at 18:44

score 1 · Answer 1 · edited Jun 01 '15 at 14:49

*tempArray[i] is the first element of the i'th vec3.

Operator precedence makes that it is evaluated *(tempArray[i]) , i.e. tempArray[i][0]. Only the first element of each vec3 is touched. If vec3 is a plain array in disguise you'll have to iterate through its elements in an inner loop.

Example code, provided typedef float vec3[3];:

for(int i=0; i<len; i++){
   for(int ii=0; ii<3; ii++){
        tempArray[i][ii] = src[i][ii];
    }
}

Btw, I would like to use this opportunity to thank iharob for his constructive comments.

Copying to a dynamically-allocated array in C

1 Answers1