Programming in C - Constants, Variables and Data Types
6. The program fragment int i = 263 ; putchar (i) ; prints
- Option : C
- Explanation : 263 in binary form is 100000111. If one tries to print an integer as a character, only the last 8 bits will be considered - the rest chopped off. So, in this case, the ASCII value of 00000111 will be printed, Look in the ASCII table. It is ringing a bell!
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7. int i = 5; is a statement in a C program.
- Option : C
- Explanation :
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8. If the declaration unsigned c:5; is replaced by unsigned: 6; then
- Option : C
- Explanation :
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9. The value of ab if ab & 0 x 3f equals 0 x 27 is
- Option : D
- Explanation : Let ab be 0 x MN. N&f should yield 7 i.e. N & 1111 should produce 0111. So, N should be 0111, i.e., 7. Similarly, M can be found to be 2. So, ab is 0 x 27.
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- Option : B
- Explanation :
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