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Asked: 2026-05-30T02:59:57+00:00

I created an union and I put different types of array inside of it.

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I created an union and I put different types of array inside of it. I printed outputs in an order and I really didn’t understand some points.

1) Why is my char array’s length always 8 even the content is different? There is only “hello” inside of it. And why does the output is “Cats rock!” when I try to print for second time. I didn’t put anything like that inside array.

2)Again length problem. Lenghts of all my arrays are 8 even the length of union. Why?

3) My last question is why double number’s value changed when I try to print for the second time.

I’m posting you my code and out put that I get. Sorry about long post, but I’m really confused.

char: hello, 8
double: 5.557111111111111, 8
int: 1937006915 1668248096 8555, 8

char: Cats rock!
double: 0.000000000000000
int: 1937006915 1668248096 8555
size of union: 8

my code

#define NUM1 5.557111111111111

#define NUM2 1937006915
#define NUM3 1668248096
#define NUM4 8555
#include <stdio.h>

/*created an union*/
typedef union {
    char  * array1;
    double num;
    int * array2;
} myunion;

int main(int argc, char ** argv)
{
    /* declaring union variable */
    myunion uni;
    /* creating my string */
    char strarray[] = "hello";

    /* declare an int array*/
    int numarray[] = { NUM2, NUM3, NUM4 };

    /* assign the string to the char pointer in the union */
    uni.array1 = strarray;

    /*  print the union and the sizeof of the pointer */
    printf("char: %s, %zu\n", uni.array1,sizeof(uni.array1));

    /* assign NUM1 to the double part of union */
    uni.num = NUM1;

    /* print the double and its sizeof */
    printf("double: %10.15f, %zu\n", uni.num, sizeof(uni.num));

    /* assign array2 of union  to the numarray */
    uni.array2 = numarray;

    /* print the values and the sizeof */
    printf("int: %d %d %d, %zu\n", uni.array2[0], uni.array2[1], uni.array2[2], sizeof(uni.array2));

    /* print the char array, double and int array */
    printf("\nchar: %s \ndouble: %10.15f \nint: %d %d %d\n",uni.array1, uni.num, uni.array2[0], uni.array2[1], uni.array2[2]);

    /* print the size of the union */
    printf("size of union: %zu\n", sizeof(uni));

    return 0;
}
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1 Answer

  1. Editorial Team

    sizeof(uni.array1) is always 8 on 64-bit platforms and 4 on 32-bit ones because it is taking the size of a pointer, not knowing how much data you believe might be behind that pointer. Similar for the arrays. C pointers which you make point to an array are “dumb” and do not understand how large the array is–you need to pass that information around separately.

    This covers your question parts 1 and 2. We like to answer specific questions here, so feel free to move your third inquiry to a separate post.

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