CODEWITHSUNDEEP
c
J. Doe
J. Doe

Reputation: 3

Logical expressions in C

Why is the result "3: 1 0 0 4", when we have incremented x? Why isn't it "3: 1 1 0 4"?

   x=0;y=4;z=3;
   printf("3: %d  %d %d %d\n", ++x || !y, x&&y, !z, y);

Upvotes: 0

Views: 58

Answers (3)

Steztric
Steztric

Reputation: 2942

As @Sami mentioned in his comment, the order that the arguments gets evaluated is compiler-specific and probably depends on calling convention. In your case the x && y argument gets computed first before the ++x || !y.

Upvotes: 0

chux
chux

Reputation: 153338

printf("3: %d  %d %d %d\n", ++x || !y, x&&y, !z, y);

++x is evaluated before !y because of the ||. !y is only evaluated if ++x result was 0.

But there is no specified order to ++x || !y vs. x&&y evaluation. Code lacks a sequence point.

Upvotes: 3

Eli Sadoff
Eli Sadoff

Reputation: 7308

If you compile this with cc you'll get the warning:

warning: unsequenced modification and access to 'x' [-Wunsequenced]`

Modifying a variable and accessing it elsewhere within printf is undefined behavior and will not give a logical result, nor the same result on every compiler.

Upvotes: 1

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