C Programming - Expressions - Discussion

Hi @Sumath,

Since C uses left associativity for evaluating expressions to break a tie between two operators having the same precedence.

In given equation, z = x + y * z / 4 % 2 - 1.

* and / has same precedence so 1st divide it into two parts like z= x + y * z and z= z / 4 % 2 - 1. Then, apply the BODMAS rule in these two respectively.

You will get * first then + from the 1st equation.
Then,

/, % and -from 2nd equation.

Then,
Add = sign at the last.

So, finally, you will get;
*/%+-=
as per their precedence.

Hope this helps.

I didn't get it properly, please explain me anyone.

How left associativity is calculated, I didn't understand? Please explain.

Order of evaluation will be compiler specific.
The following program gave me output b = 35..
#include<stdio.h>

int main()
{
int a = 10, b;
b = a++ + ++a + ++a;
printf("b = %d\n",b);
return 0;
}

BUT the GCC also throws a warning as follows:

gcc -Wall -o "scrapbook" "scrapbook.c" -lm (in directory: /home/crownedeagle/Downloads/CDAC C-CAT/C - K&R Solutions)
scrapbook.c: In function \'main\':
scrapbook.c:6:15: warning: operation on \'a\' may be undefined [-Wsequence-point]
b = a++ + ++a + ++a;
^
scrapbook.c:6:15: warning: operation on \'a\' may be undefined [-Wsequence-point]
Compilation finished successfully.

So the compilation of program and evaluation of 'b' will be undefined, it may give different answers on different machines and on different compiler.

My advice: Don't get stuck on this too much.

@Mauli.
For your program.

The execution like this,
1) b+=++a which is equal to b=b+ ++a i.e b=20+11=31(stored in memory b=31 and a=11)
2) ++b=32( stored in memory).
3) a++=11( a=12 stored in memory now).
4) a+=b++ which is equal to a=a+ b++ i.e a=12+ 32=44( stored in memory a=44 and b=33 Because b++ is 1st assign then store in memory).

So the output is b=33 and a=44.

In don't understand the answer can anyone please explain it clearly?

Hi, I am not getting this, Please anyone explain me.

Int a=1 ,b=2,c=3
Switch (p)
{
Case1:a++
Case2:++b
Break;
Case 3:c--;
}
System.out.println(a+","+b+","+c);

When 1:p=1 2:p=3.

#include<stdio.h>
main()
{
int a=10,b=20;
b+=++a;++b;a++;
a+=b++;
printf("a=%d \n b=%d",a,b);
}
a=44
b=33

Please explain this program.
How b=33 and a=44?

#include<stdio.h>
main()
{
int x=5,y=6;
x=++y;
y++=x--;
x++=6+(++y);
printf("%d%d\n",x++,y);
}

Anyone explain it, please.