Why isn't the output right?

Question

union test
{
int x;
char arr[4];
int y;
};

#include<stdio.h>
int main()
    {
    union test t;
    t.x = 0;
    t.arr[1] = 'G';
    printf("%s", t.arr);
    return 0;
    }

Predict the output of the above program. Assume that the size of an integer is 4 bytes and size of a character is 1 byte.

Why shouldn't the answer be Garbage character followed by 'G', followed by more garbage characters?

`t.x = 0` sets all 4 characters to 0. Do you know how unions work? — HolyBlackCat, Jan 24 '21 at 14:13
("size of character is 1 byte" - C guarantees `sizeof(char)==1` _always_) — aschepler, Jan 24 '21 at 14:26
I mean first character of array has been initialised as G and others should contain garbage value. @klutt — Girik Garg, Jan 24 '21 at 14:35
@HolyBlackCat With utmost humility, Can u justify your claim? — Girik Garg, Jan 24 '21 at 14:36
@GirikGarg This is very basic knowledge. Any site explaining unions should be able to explain this. — klutt, Jan 24 '21 at 14:40
Note that arrays are indexed from 0, not 1. The assignment `t.arr[1] = 'G';` is not initializing the first element of the string. — Jonathan Leffler, Jan 24 '21 at 14:49

score 1 · Accepted Answer · answered Jan 24 '21 at 14:47

Assuming 4 byte ints here:

t.x = 0;

write four 0 bytes into the union. That overwrites all the elements in arr[].

t.arr[1] = 'G';

writes a 'G' into the second byte of the union. There are still 3 zero-bytes: one before the G and two after the G.

printf("%s", t.arr);

prints up to the first 0 byte in t.arr, which is the first byte in t.arr because you wrote 4 zeroes into the union. No output.

Why isn't the output right?

1 Answers1