C union char array prints 'd' on mac? - c++

Question

I am new to C/C++, but was curious to know about the issue i am seeing.
typedef union
{
int a;
float c;
char b[20];
}
Union;
int main()
{
Union y = {100};
printf("Union y :%d - %s - %f \n",y.a,y.b,y.c);
}
And output is
Union y :100 - d - 0.000000
My question is ...why is 'd' getting printed? I changed the order in union still the same output. but if i declare a char f[20] outside the union then nothing gets printed.
I am having MacBook lion image and using xcode.
THanks in advance

Answer 1

The ASCII code for 'd' is 100. Setting a to 100 amounts to setting b to {'d', '\0', '\0', '\0', …noise…} (on a 32-bit little-endian machine), which printf treats as "d".

Answer 2

The following program may help you understand better:
#include <stdio.h>
typedef union
{
int a;
float c;
char b[20];
}
Union;
void dump(const void* buffer, size_t length)
{
size_t i;
for (i = 0; i < length;) {
printf("%.2x ", reinterpret_cast<const unsigned char*>(buffer)[i]);
++i;
if (i % 16 == 0) {
putchar('\n');
} else if (i % 8 == 0) {
putchar(' ');
}
}
if (i % 16 != 0) {
putchar('\n');
}
}
int main()
{
Union y = {100};
printf("Union y :%d - %s - %f \n",y.a,y.b,y.c);
printf("The content of the Union is: \n");
dump(&y, sizeof y);
}
The output is:
Union y :100 - d - 0.000000
The content of the Union is:
64 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00
Effectively, the binary representation of a (which is a int) is 64 00 00 00, 100 in little-endian hexadecimal. The binary representation of b is 64 00 ..., and the 0x00 ends the string, while 0x64 is 'd'. The binary representation of c is 64 00 00 00, which in the IEEE float representation is 0.0, because the non-zero part is the exponent.

Answer 3

I changed the order in union still the same output.
The order of the elements in a union doesn't change anything because all the elements of a union use the same piece of memory. Your code prints 100 for y.a and d for y.b because both expressions interpret the same bytes. So, for example, if you add a line that sets y.b and then prints again:
Union y = {100};
printf("Union y :%d - %s - %f \n",y.a,y.b,y.c);
y.b = 'f';
printf("Union y :%d - %s - %f \n",y.a,y.b,y.c);
you'll see that y.a and y.c. change whenever y.b changes, and vice versa. y.a should change to 102 in the second printf(), since that's the ASCII character code for 'f'.