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14 C Puzzle Solution: August 2011

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14 c puzzle solutions. Solution C Puzzle #14. How about unrolling the loop: it has a fixed number of iterations,. Since i goes from 0 to 4, and shift goes through the sequence 1, 2, 4,. 8, and 16, regardless of the input. So the calculation has five fixed. Stages, which have been rolled up into a loop for brevity. These five. Stages limit it to counting up to 32 bits (2 to the power of 5). At. Each stage, the following is evaluated:. X = (x and mask[i]) ( x shift) and mask[i]);. Each stage does this:.

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14 C Puzzle Solution: solution C Puzzle 5 To 7

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14 c puzzle solutions. Solution C Puzzle 5 To 7. Solution C Puzzle #5. The # operator stringizes its operand, i.e. it puts double quotes. The # operator 'glues' its operands together. The preprocessor expands macro definitions in the text. It keeps expanding macros as long as it finds any; to block. It won't expand a macro that is being expanded already. In the first printf(), firstly, it is replacing h(f(1,2) by. G(f(1,2) , and then immediately in the same pass. In short, in first printf(). Next, the la...

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14 C Puzzle Solution: June 2011

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14 c puzzle solutions. Solution C Puzzle #5. The # operator stringizes its operand, i.e. it puts double quotes. The # operator 'glues' its operands together. The preprocessor expands macro definitions in the text. It keeps expanding macros as long as it finds any; to block. It won't expand a macro that is being expanded already. In the first printf(), firstly, it is replacing h(f(1,2) by. G(f(1,2) , and then immediately in the same pass. In place of h(f(1,2) . This g("12") results in a string "12". Files...

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14 C Puzzle Solution: solution C Puzzle 9

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14 c puzzle solutions. Solution C Puzzle 9. Solution C Puzzle #9. This is an excellent question. This is not a bug but a fundamental property of floating point numbers - they are not exact representations of numbers. Floating point numbers are represented by an exponent multiplied by an arithmetic series.There are gaps between floating point numbers where the number suffers from what is called a floating point approximation error. Printf("%.20f n" , f);. 05 for example does not fall between a gap so.

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14 C Puzzle Solution: solution C Puzzle 8

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14 c puzzle solutions. Solution C Puzzle 8. Solution C Puzzle #8. It has a fixed number of iterations,. Since i goes from 0 to 4, and shift goes through the sequence 1, 2, 4,. 8, and 16, regardless of the input. So the calculation has five fixed. Stages, which have been rolled up into a loop for brevity. These five. Stages limit it to counting up to 32 bits (2 to the power of 5). At. Each stage, the following is evaluated:. X = (x and mask[i]) ( x shift) and mask[i]);. Each stage does this:. X = F i(x).

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14 C Puzzle Solution: solution C Puzzle 3 and 4

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14 c puzzle solutions. Solution C Puzzle 3 and 4. Solution C Puzzle #3. Answer: This will print "1" only once. Clear your concepts regarding continue statement. Courtsey: msdn.microsoft.com. Statement passes control to the next iteration of the nearest enclosing do. Statement in which it appears, bypassing any remaining statements in the do. The next iteration of a do. Statement is determined as follows:. Statement, the next iteration starts by reevaluating the expression of the do. Statement in a for.

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14 C Puzzle Solution: solution C Puzzle 10 To 13

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14 c puzzle solutions. Solution C Puzzle 10 To 13. Solution C Puzzle #10. Puts ("Hello, world! Because i have told you not to use 'a' semicolon. So this could also be the answer. Solution C Puzzle #11. Error in the line int. Here the declaration and definition is happening at the same time and hence compiler. Associates/initializes a to 1 but for 2 it considers as variable and since it. Is a not a valid variable(variable names cannot start with digits) it throws. Solution C Puzzle #12. Dst ) = *(src );.

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14 C Puzzle Solution: solution C Puzzle #14

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14 c puzzle solutions. Solution C Puzzle #14. How about unrolling the loop: it has a fixed number of iterations,. Since i goes from 0 to 4, and shift goes through the sequence 1, 2, 4,. 8, and 16, regardless of the input. So the calculation has five fixed. Stages, which have been rolled up into a loop for brevity. These five. Stages limit it to counting up to 32 bits (2 to the power of 5). At. Each stage, the following is evaluated:. X = (x and mask[i]) ( x shift) and mask[i]);. Each stage does this:.