why the result of print statement for a.x isn''t 1?
resultCode:#include<stdio.h> union u{ int x; char y;}a; int main () { a.x = 1; a.y = 'A'; printf ("%d ", a.x); printf ("%c ", a.y); printf ("\n %d ", sizeof(a.x)); printf ("\n %d ", sizeof(a.y)); return 0; }
65 A
4
1
65 A
4
1
Last edited by skyr6546; 04-03-2020 at 09:57 PM.
> why the result of print statement for a.x isn''t 1?
Because you have a union and not a struct.
With a union, only the last member you wrote is valid to read from.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Because that is how union work on your computer because of byte ordering of the CPU.
Tim S.
"...a computer is a stupid machine with the ability to do incredibly smart things, while computer programmers are smart people with the ability to do incredibly stupid things. They are,in short, a perfect match.." Bill Bryson
why getting wired resultOriginally Posted by Salem
1092721050 � 10.100000Code:#include<stdio.h>union u { int x; char y; float z; }a; int main () { a.x = 1; a.y = 'A'; a.z = 10.10; printf ("%d ", a.x); printf ("%c ", a.y); printf ("%f ", a.z); return 0; }
...Program finished with exit code 0
Press ENTER to exit console.
Did you read the replies you were given? They answer your question.
If you're not interested in reading what people tell you about unions, then perhaps you shouldn't bother with unions. Change the union to struct and the "weirdness" will vanish.
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