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Answer to Question #53572 in C for rajashree g.s.

Question #53572
int main()
int x=15;
printf("\n %d %d %d",x!=15,x=20,x<30);
return 0;

In this program x is already assigned the value 15. what will be the output and explain sir.
Expert's answer
The output of the program is not defined by standard. Because output of this program depends on the order in which the function arguments are evaluated. In this case “,” is a separator of function arguments and this is not a sequence point.

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