This question already has answers here:
Uninitialized variable in C [duplicate]
(6 answers)
Closed 7 years ago.
Recently I have installed Ubuntu and obviously I am compiling my C code in gcc. I came across the following code :
#include <stdio.h>
main()
{
int i = 10,j = 20, k;
printf("i=%d j=%d k=%d\n", i, j, k);
}
The output is coming as ::
i=10 j=20 k=0
But as far as I know that the output for the value of k should be a Garbage value since it has not been initialized.
Is there something that I am missing here?
You cannot expect anything from the program you posted, 0 is a perfectly possible garbage value.
Also, main() must have a return value and it must be int.
It's undefined behavior, so anything is possible, including 0 to be printed.
How is 0 not a garbage value?
Related
This question already has answers here:
Initializing variables in C
(10 answers)
Closed 6 years ago.
#include <stdio.h>
#include <string.h>
#include "prac.h"
#define MYNAME "Butter"
int main() {
int numberOfKids;
int weight;
int shirt;
printf("If I eat a Watermelon I will weigh %d lbs \n", weight + numberOfKids+ shirt );
return 0;
}
I compiled and ran the program and the result was 1; although I expected it to be 0. When I checked the value of each variable individually, the weight variable's value was 1. Can someone explain why that specific variables result was not 0? I am new to C and want to experiment with the basics to get a deeper understanding of the nuances of C. Any help would be appreciated.
Variables inside a function in C are not guaranteed to be set to anything by default. In memory, whatever was last stored there (which might not be flushed/erased to be 0) will be what the int is initialized to.
This is answered in Initializing variables in C
EDIT: As chux has stated below, local static variables are initialized to 0 if they aren't given an initial value. Also covered Is un-initialized integer always default to 0 in c?
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 6 years ago.
I am trying the following code:
#include <stdio.h>
int main()
{
int c =0;
c -= --c - c++;
printf("%d \n",c);
return 0;
}
When I compiled and run it using a online c complier (https://www.tutorialspoint.com/compile_c_online.php) the result is -1. But I expected it to be 0.
So, I try it on my local Dev C++ (Windows) and the result is 0.
Should the result be 0 ?
If so, why 2 gcc compilers (ok they are in different plataform) gives me 2 different results ?
I ve been looking for some kind of automatic flag otimization which could produce a different result but I had no success.
THIS IS UNDEFINED BEHAVIOR (3 modifications without sequence points inbetween to the same variable)
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 9 years ago.
I felt that such an expression should be invalid but I was able to compile it and got the answer 5.
At the end I felt that even if it does answer should be 4 not 5.
int main(void)
{
int i=1;
// how is the next line evaluated ie in what sequence??
i=2+2*i++;
printf("%d",i);
return 0;
}
The output I got was 5. I can not understand how it should give the value.
This is undefined behaviour, since i is modified more than once between sequence points. For instance, this compiler gives 4 as the answer, because it puts the increment after the assignment. Another reasonable answer is 6, if the increment is before the assignment. But, as you've found, the compiler is permitted to make the answer whatever it wants, including 5.
See here for more about sequence points and undefined behaviour.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
There is a code following below and i ma facing a very serious problem in understanding the logic for the code.
#include <stdio.h>
#include <stdlib.h>
int main(void )
{
int i = 1 ;
printf("\n%d %d %d %d\n",++i,i++,i++,++i) ;
return 0 ;
}
I am using gcc compiler under the linux distro named Mandriva. In the above mentioned i have used pre and post increment with a variable in the printf statement.
The output that i am supposed to get is 2 2 3 5, but i am getting a different output.
Please help me in this code.
I am feeling much difficult in this code.
It's undefined behavior. There's no sequence points between the increments of i.
Any result is a correct result (including your hard drive being formatted).
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
int main()
{
int a=5,s;
s=++a + ++a;
printf("%d",a);
printf("%d",s);
}
output is 7 and 14
BUT
int main()
{
int a, s;
printf("Enter value of a");
scanf ("%d",&a);
s=++a + ++a;
printf("%d",a);
printf("%d",s);
}
input user gives is 5
output is 7 and 13
WHY?
Undefined behaviour:
s=++a + ++a;
Anything can happen when undefined, so your behaviour is perfectly valid.
I'd suspect this is an artifact of compiler optimisation, in the first example a is known so the compiler optimises the preincrements to occur before the addition. In the second example the value is unknown and the compiler does not optimise the sequence causing it to complete left to right. This may be a function of your specific compiler and it would need to be looked at specifically.
Undefined behaviour. Change it, or you risk being attacked by raptors.
hi budy this coding working correctly in VI compiler ..