Double array only stores ints

Question

I have this code here that I think should work, and it does! except for the fact that though I've declared a pointer to an array of type double, it always stores int, and I don't know why.

first, I have this struct and it is defined like this:

struct thing {
    int part1;
    double *part2;
};

then I initialize the thing by saying struct thing *abc = malloc (sizeof(struct thing)) and part1 = 0 and part2 = malloc(sizeof(double)) .

then, I try to set specific values at specific positions in the array of double. This works fine with integers, but when I tried 0.5, it set the value to 0. when I tried 2.9, it set the value to 2. I really don't know why it does this. code for setValue looks like this:

struct thing *setValue (struct thing *t, int pos, double set){
    if (t->part1 < pos){ // check to see if array is large enough
        t->part2 = realloc (t->part2, (pos+1) * sizeof(double));
        for (int a = t->part1 + 1; a < pos + 1; a++)
            t->part2[a] = 0;
    t->part1 = pos;
    }
    t->part2[pos] = set; // ALWAYS stores an integer, don't know why
    return t;
}

-- Edit: So there is nothing really mallicious about this part; but here's the rest of my code JUST IN CASE:

Relevant functions that operate on my struct thing

#include "thing.h"
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

struct thing *makeThing(){ // GOOD
    struct thing *t = (struct thing *) malloc (sizeof(struct thing));
    t->part1 = 0;
    t->part2 = malloc (sizeof(double));
    t->part2[0] = 0;
    return t;
}

struct thing *setValue (struct thing *t, int pos, double set){
    if (t->part1 < pos){ // check to see if array is large enough
        t->part2 = realloc (t->part2, (pos+1) * sizeof(double));
        for (int a = t->part1 + 1; a < pos + 1; a++)
            t->part2[a] = 0;
    t->part1 = pos;
    }
    t->part2[pos] = set; // ALWAYS stores an integer, don't know why
    return t;
}

double getValue (struct thing *t, int pos){
    if (pos <= t->part1){
        return t->part2[pos];
    }
    return 0;
}

Header file:

#ifndef THING_H
#define THING_H

struct thing {
    int part1;
    double *part2;
};

struct thing *makeThing();
struct thing *setValue (struct thing *t, int pos, double set);
double getValue (struct thing *t, int pos);

#endif

main file:

#include <stdio.h>
#include "thing.h"

int main (void)
{
    struct thing *t = makeThing();
    setValue (t, 1, -1);
    setValue (t, 1, -2);
    setValue (t, 10, 1);
    setValue (t, 3, 1.5);

    printf ("%g\n", getValue (t, 0));
    printf ("%g\n", getValue (t, 1));
    printf ("%g\n", getValue (t, 2));
    printf ("%g\n", getValue (t, 3));
    printf ("%g\n", getValue (t, 4));
    printf ("%g\n", getValue (t, 5));
    printf ("%g\n", getValue (t, 6));
    printf ("%g\n", getValue (t, 7));
    printf ("%g\n", getValue (t, 8));
    printf ("%g\n", getValue (t, 9));
    printf ("%g\n", getValue (t, 10));

    return 0;
}

On my computer, this prints out: 0 -2 0 1 0 0 0 0 0 0 1

EDIT: Turns out that when I compile it via codeblocks, it works...

Ultimately, I am confused.

Answer 1

Double converts to int in C?

No, it doesn't. When you assign a double value to an object ot type double there is no conversion whatsoever.

Your problem is not in the code you've shown; it is somewhere else (the way you print, some stupid #define , some other thing).

Oh! And you really should make sure the realloc() s work. Otherwise, instead of an error, users may get a slightly erroneous value...

As I said in a comment, your code works as expected at ideone .

Answer 2

I think you might have messed up your format specifier in your printing. Be sure to have the correct specifiers for the correct data types.

I know that happened to me sometimes when I changed data types in my Objective-C code. :)