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20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:??????

Answers were Sorted based on User's Feedback



20. main() { int i=5; printf("%d%d%d%d%d%d",i++,i--,++i,--i,i); } Answer:??????..

Answer / ashok kumar

45545

Is This Answer Correct ?    6 Yes 0 No

20. main() { int i=5; printf("%d%d%d%d%d%d",i++,i--,++i,--i,i); } Answer:??????..

Answer / ashish

The only concept here is that the printf starts 'generating'
results from RHS to LHS.
Hence, in the process of generation: i=5; then --i=4; ++i=5;
i--=5(due to the postfix concept, but after finishing this
calculation i=4); now i++=4(after performing i++, the value
of 'i' is 5, but for printing, due to the postfix concept,
i=4, the current value of 'i' will be printed).

The printing happens in order from LHS to RHS. Hence these
values are printed as 4 5 5 4 5.

Is This Answer Correct ?    5 Yes 1 No

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