Nested loop assignment to struct member

Question

I'd like to have an array of instances of TestStruct with a single member of type int. I'd like the output to be

1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6

but it is 0, 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

I'm a little stumped, because I don't really get how this isn't working:

#include <stdio.h>
#include <stdlib.h>

typedef struct {
    int num;
} TestStruct;

int main(void) {
    int i, j;
    TestStruct *test = malloc(sizeof(TestStruct) * 18);

    for (i = 1; i <= 3; i++) {
        for (j = 0; j < 6; j++) {
            test[i].num = i;
        }
    }
    for (i = 0; i < 18; i++) {
        printf("%d ", test[i].num);
    }

    return 0;
}

If I just call printf() inside a nested for-loop, it works fine:

#include <stdio.h>

int main(void) {
    int i, j;

    for (i = 1; i <= 6; i++) {
        for (j = 0; j < 3; j++) {
            printf("%d ", i);
        }
    }
}

Why is this happening?

Your printing loop that "works fine" has the indices swapped if you compare it to the desired output. — M Oehm
– M Oehm, Commented Dec 1, 2015 at 20:32
@MOehm Fixed. I did a lot of edits and compiling, and apparently I got them swapped. Thank you! :) — jmkjaer
– jmkjaer, Commented Dec 1, 2015 at 20:42

Afshin · Accepted Answer · 2015-12-01 20:29:44Z

3

you need to write it like this:

for (i = 0; i < 6; i++) {
        for (j = 0; j < 3; j++) {
            test[i*3+j].num = i+1;
        }
  }

answered Dec 1, 2015 at 20:29

Afshin

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1 Comment

jmkjaer Over a year ago

That makes sense. Accepted answer. Thank you!

Nakib · Accepted Answer · 2015-12-01 20:37:22Z

2

Change you code to this, You should be indexing i element not j

 for (i = 1; i <= 3; i++) {
        for (j = 0; j < 6; j++) {
            test[j + (3*i)].num = i;
        }
  }

The above code will print 111111222222333333

In order to achieve 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6 your code should look like below

for (i = 1; i <= 6; i++) {
            for (j = 0; j < 3; j++) {
                test[j + (3*i)].num = i;
            }
 }

Update: You need to keep track of outer loop or index you are appending number to so it does not rewrite everytime outer loop is run. I did it like this test[i + (3*i)]

edited Dec 1, 2015 at 20:37

answered Dec 1, 2015 at 20:27

Nakib

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7 Comments

M Oehm Over a year ago

This will initialise six instead of three elements. You need a combination of indices or a third, independent index.

Afshin Over a year ago

Nabik, 2nd one has same problem again. you only write from element 1..6 in array. Result will be 012345600000000000

Afshin Over a year ago

LOL, code still has problem. result will be 'segmentation fault'. :D

jmkjaer Over a year ago

@Nakib I'll have to fiddle with it for a bit tomorrow. I accepted Afshin's post as answer, but thank you too!

M Oehm Over a year ago

Nakib, please take some care when posting your answers. Ideally, test them before posting them. You update repeatedly, but the updates aren't much better. For example, j + (3*i) will have values of 18 through 20 in the last iteration of the outer loop. These indices are out of bounds.

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