CODEWITHSUNDEEP
cassignment-operator
Heatblast
Heatblast

Reputation: 65

assignment operators and c

#include <stdio.h>

int main()
{
   int i,j;
   i=j=(22,23,24);
   printf("i:%d",i);
   printf("\nj:%d",j);
}

this is giving output of both i,j as 24.

#include <stdio.h>

int main()
{
   int i,j;
   i=j=22,23,24;
   printf("i:%d",i);
   printf("\nj:%d",j);
}

and this gives both i,j as 22. Can someone explain the terminlology behind. TIA

Upvotes: 2

Views: 93

Answers (2)

user15943018
user15943018

Reputation:

A comma separated list of constant expressions does not make much sense. The main purpose of comma operator is to separate assignments - a transparent and conscious side-effect.

With parentheses and two more assignments it groups as:

i = j = (22,x=23,y=24);

Since an expression in parens (E) is a primary expression, this can be hierarchically seen as

i = j = E;

This E consists of three comma separated (sub-)expressions where the last one counts. This is assignment y=24 with result/evaluation 24.

The 22 is thrown away, but x=23 has the obvious side effect.

Normally E would be something like 2*n+1 that does not need parens. Until up to the i = j > 22 ? 23 : 24 (where it gets a bit confusing).

The second one groups:

i=j=22, x=23, y=24;

This starts to make sense now. It is just three assigments on one line. 23 and 24 are not "lost".

The assigment expression i=j=22 itself has one level of recursion; it works right-to-left, which is only natural.

But the examples are more about combining separating comma operators , and grouping parentheses ( ).

Upvotes: 0

MikeCAT
MikeCAT

Reputation: 75062

An assignment operator = has higher precedence than a comma operator ,.

C Operator Precedence - cppreference.com

In this statement

i=j=(22,23,24);

(22,23,24) is firstly calculated. The 22 and 23 are ignored by the comma operator and it is evaluated to 24. Then, the result 24 is assigned to j, and the value is also assigned to i.

On the other hand, in this statement

i=j=22,23,24;

i=j=22 is firstly calculated. This assigns 22 to j, then assign the value to i. After that, the evaluation result of assignment operator 22, and an integer literal 23 are ignored by the comma operator. Finally the expression is evaluated to the value 24, which is also ignored.

Upvotes: 5

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