c - Explaination for printf with comparing variables as arguments -

main(){     int = 5;     int b = 6;     printf("%d %d %d",a==b,a=b,a<b); } 

output in testing

1 6 1

in above program expecting output 0 6 0 . in compilers giving output (e.g. xcode) in other compilers giving output 1 6 1 . couldn't find explanation . case of sequence point.

consider below program

main(){     int = 5;     int b = 6;     printf("%d %d %d",a<b,a>b,a=b);     printf("%d %d",a<=b,a!=b); } 

output in testing

0 0 6 1 0

this below program giving correct output expecting 0 0 6 1 0 why above program not giving output 060 in of compilers

c standard says:

c11: 6.5 (p2):

if side effect on scalar object unsequenced relative to either different side effect on same scalar object or a value computation using value of same scalar object, behavior undefined [...]

this means program invokes undefined behavior. in statements

printf("%d %d %d",a==b,a=b,a<b);   

and

printf("%d %d %d",a<b,a>b,a=b); 

the side effect a unsequenced because standard says:

6.5.2.2 (p10):

[...] every evaluation in calling function (including other function calls) not otherwise sequenced before or after execution of body of called function indeterminately sequenced respect execution of called function.