precedence and order of evaluation

Operators Precedence in C

Operator precedence determines the grouping of terms in an expression and decides how an expression is evaluated. Certain operators have higher precedence than others; for example, the multiplication operator has a higher precedence than the addition operator.
For example, x = 7 + 3 * 2; here, x is assigned 13, not 20 because operator * has a higher precedence than +, so it first gets multiplied with 3*2 and then adds into 7.
Here, operators with the highest precedence appear at the top of the table, those with the lowest appear at the bottom. Within an expression, higher precedence operators will be evaluated first.
CategoryOperatorAssociativity
Postfix() [] -> . ++ - -Left to right
Unary+ - ! ~ ++ - - (type)* & sizeofRight to left
Multiplicative* / %Left to right
Additive+ -Left to right
Shift<< >>Left to right
Relational< <= > >=Left to right
Equality== !=Left to right
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

Example

Try the following example to understand operator precedence in C −
#include <stdio.h>

main() {

   int a = 20;
   int b = 10;
   int c = 15;
   int d = 5;
   int e;
 
   e = (a + b) * c / d;      // ( 30 * 15 ) / 5
   printf("Value of (a + b) * c / d is : %d\n",  e );

   e = ((a + b) * c) / d;    // (30 * 15 ) / 5
   printf("Value of ((a + b) * c) / d is  : %d\n" ,  e );

   e = (a + b) * (c / d);   // (30) * (15/5)
   printf("Value of (a + b) * (c / d) is  : %d\n",  e );

   e = a + (b * c) / d;     //  20 + (150/5)
   printf("Value of a + (b * c) / d is  : %d\n" ,  e );
  
   return 0;
}
When you compile and execute the above program, it produces the following result −
Value of (a + b) * c / d is : 90
Value of ((a + b) * c) / d is  : 90
Value of (a + b) * (c / d) is  : 90
Value of a + (b * c) / d is  : 50

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