CODEWITHSUNDEEP
cvariables
Carlo Di Dato
Carlo Di Dato

Reputation: 51

Why gcc on Windows assigns a specific value to a variable?

this is my first question so... nice to meet you! Could someone explain why this code written in C

#include <stdio.h>

int main(){
 int choice;

 printf("\nSize of 'choice' %d\n", sizeof(choice));
 printf("Size of 'int' %d\n", sizeof(int));
 printf("Value of 'choice %d\n", choice);
 return 0;
}

compiled on Windows using gcc -o C:\test.exe test.c (gcc version 2015/06/27) returns these values

Size of 'choice' 4
Size of 'int' 4
Value of 'choice 2

while on Linux everything works as I expected, returning these values

Size of 'choice' 4
Size of 'int' 4
Value of 'choice 0

What am I missing?

Thanks a lot.

Upvotes: 2

Views: 79

Answers (5)

Anjaneyulu
Anjaneyulu

Reputation: 434

The value in the choice variable is not initialized so default value is undefined since it is a automatic storage class and it is different in different platforms

Upvotes: -1

Alexander Paschenko
Alexander Paschenko

Reputation: 741

You try to use variable without initialization. So, it can contains a random value

Upvotes: 1

Mr.C64
Mr.C64

Reputation: 42984

In your code you defined a variable without specifying an initial value, so you cannot pretend the variable to be automatically initialized to anything.

If you want your variable to be zero-initialized, please do so explicitly with =0.

Upvotes: 3

tdao
tdao

Reputation: 17688

What am I missing?

You never initialise choice in the first place. So the behaviour is undefined. It doesn't matter what compiler you are using, the value printed is just garbage value in the memory location of choice.

Upvotes: 2

Some programmer dude
Some programmer dude

Reputation: 409404

Uninitialized local non-static variables are just that, uninitialized. They will have an indeterminate value. Using them without initialization leads to undefined behavior.

Upvotes: 7

Related Questions