# include<stdio.h>
void main()
{ int p;
p = (1,2,2,100,1);
printf("%d",p);
}
I dont know what will be the output . Can anyone please tell me what exactly second line implies ?? :rolleyes:
# include<stdio.h>
void main()
{ int p;
p = (1,2,2,100,1);
printf("%d",p);
}
I dont know what will be the output . Can anyone please tell me what exactly second line implies ?? :rolleyes:
comma ' , ' operator has left to right associativity
# include<stdio.h>
void main()
{ int p;
p = 1,2;
printf("%d",p);
}
In this case output will be 1
But if we apply ( )
then last value is assigned to p and output will be 2
# include<stdio.h> void main() { int p; p = (1,2,2,100,1); printf("%d",p); }
I dont know what will be the output . Can anyone please tell me what exactly second line implies ?? :rolleyes:
though p should be declare as an array int p[]
if p were an array, then the initializer list would have to be in braces, not parentheses, like this
int p[] = {1,2,2,100};
But the above is entirely different than the code you originally posted.
># include<stdio.h>
>void main()
>{ int p;
> p = 1,2;
> printf("%d",p);
>}
>In this case output will be 1
what will be the output, if i make a bit of change in the existing code, i mean:::
# include<stdio.h>
void main()
{ int p = 1,2;
printf("%d",p);
}
i got error in this case, i dont know , why its producing error in this case, when it looks similar to the old one?? :?:
parentheses are required as shown in previous posts.
The same..
Paranthesis will do the work for getting p=(1,2)
so o/p will be 2.
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