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Ans. (d) double –5.0
Explanation:
The execution goes on like this:
– 4 + 1/2 + 2*-3 + 5.0;
– 4 + 0 + -6 + 5.0; // integer division: 1/2 truncates .5
– 10 + 5.0; // higher type is double 5.0, so -10 is casted to double
– 5.0; // finally, double -5.0.
Ans. (c) 1
Explanation:
It is printed one time only because break; will terminate the current loop.
Explanation:
z = ++x * (y– –)– y
= 21 * (10) – 9
= 210 – 9 = 201
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Download NowChapter No. | Chapter Name |
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Chapter 1 | Revision of Class IX Syllabus |
Chapter 2 | Class as a Basis of all Computation |
Chapter 3 | User - defined Methods |
Chapter 4 | Constructors |
Chapter 5 | Library classes |
Chapter 6 | Encapsulation |
Chapter 7 | Arrays |
Chapter 8 | String handling |
Chapter Wise Important Questions for ICSE Board Class 10 Computer Applications |
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Revision of Class IX Syllabus |
Class as a Basis of all Computation |
User - defined Methods |
Constructors |
Library classes |
Encapsulation |
Arrays |
String handling |
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