It dictates the order of evaluation of operators in an expression.
If two operators of same precedence (priority) is present in an expression, Associativity of operators indicate the order in which they execute.
| Category | Operator | Associativity |
|---|---|---|
| Postfix | () [] -> . ++ - - | Left to right |
| Unary | + - ! ~ ++ - - (type)* & sizeof | Right to left |
| Multiplicative | * / % | Left to right |
| Additive | + - | Left to right |
| Shift | << >> | Left to right |
| Relational | < <= > >= | Left to right |
| Equality | == != | Left to right |
| Category | Operator | Associativity |
|---|---|---|
| Bitwise AND | & | Left to right |
| Bitwise XOR | ^ | Left to right |
| Bitwise OR | | | Left to right |
| Logical AND | && | Left to right |
| Logical OR | || | Left to right |
| Conditional | ?: | Right to left |
| Assignment | = += -= *= /= %=>>= <<= &= ^= |= | Right to left |
| Comma | , | Left to right |
#1 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int a = 4;
printf("\n %d", 10 + a++);
printf("\n %d", 10 + ++a);
return 0;
}
Output:
14
16
#2 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int a=4, b=5, c=6;
a = b == c;
printf("\n a = %d",a);
return 0;
}
Output:
a = 0
#3 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int a=1, b=2, c=3, d=4, e=5, res;
res = a + b/c -d * e;
printf("\n Result = %d", res);
res = (a + b) / c - d * e;
printf("\n Result = %d", res);
res = a + (b/(c-d)) * e;
printf("\n Result = %d",res);
return 0;
}
Output:
Result = -19
Result = -19
Result = -9
#4 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int a=1,b=2,c=3;
printf("++a - b++ - --c - --c + %d \n",a);
printf("--a + %d \n",++b - c + --a);
return 0;
}
Output:
++a - b++ - --c - --c + 1
--a + 0
#5 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int a=100,b=3;
float c;
c = a/b;
printf("\n c = %f",c);
c=(a*++b)/3.0;
printf("\n c = %d",(int)c);
return 0;
}
Output:
c = 33.000000
c = 133
#6 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int a=1,b=3,c=6,ans=0;
printf("\n %d", (a>b && a<c));
ans *= --b + ++a*c++;
printf("\n %d",ans);
printf("\n c = %d",(b>10 || (b + a)>0 || a != b));
return 0;
}
Output:
0
0
c = 1
#7 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int x=2,y=4,z=1,ans;
z += y++ - ++x;
printf("\n %d", z);
ans = ++x + ++x * --x * --z;
printf("\n %d",ans);
ans = x+++y;
printf("\n %d %d %d %d",x,y,z,ans);
return 0;
}
Output:
2
24
5 5 1 9
#8 : Find the o/p of the below program.
#include <stdio.h>
int main(){
int x=4;
char ch = 'E';
printf("\n %d", ch+x);
ch += x*8;
printf("\n %c",ch);
return 0;
}
Output:
73
e