Answer 1)
OUTPUT: 1 0 1
In the print statement we have three digits that will be printed.
For the first 10 > -1 returns true which when converted into digit gives 1. The conversion to digit occurs becuase we have used %d.
For the second (unsigned)10 > -1 returns false which when converted into digit gives 0. The conversion to digit occurs becuase we have used %d.
For the second (unsigned)(10 > -1) returns true which when converted into digit gives 1. The conversion to digit occurs becuase we have used %d.
CODE AND OUTPUT SCREENSHOT
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